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Digitized  by  the  Internet  Archive 

in  2007  with  funding  from 

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http://www.archive.org/details/elementsofplanegOOibacrich 


Buttery's  Series  of  Mathematics. 


ELEMENTS 


OP 


PLANE  GEOMETRY. 


BY 


FRANKLIN  IBACH,  B.  S., 

TEACHER  OF  MATHEMATICS  IN  THE  PEIRCE  COLLEGE  OF  BUSINESS. 


PHILADELPHIA; 
E.  H.  BUTLER   &  CO. 


0  ■ 
Z3 


Entered,  according  to  Act  of  Congress,  in  the  year  1882,  by 

E.  H.  BUTLER  &  CO. 

In  the  Office  of  the  Librarian  of  Congress,  at  Washington,  D.  C. 


ELECTROTYI'KD    BY    MACKELLAR,  SMITHS    4    JORDAN,  PHILADELPHIA. 


CONTENTS, 


INTRODUCTION. 


PAGE 


Subject-Mattek 7 

BOOK  I. 

Definitions 9 

Mathematical  Terms 14 

Axioms 14 

Postulates 15 

Demonstration 15 

Symbols  and  Abbreviations 16 

Perpendicular  and  Oblique  Lines       ....  17 

Parallel  Lines 27 

Equality  of  Angles  ....        .       .        .        .31 

Triangles 34 

Relation  between  the  Sides  of  a  Triangle  .        .  36 

Medial  Lines 38 

Angles  of  a  Triangle     .......  39 

Equality   of  Triangles 41 

Relation  between  the  Parts  of  a  Triangle     .        .  48 

Bisectors  of  Angles 51 

Polygons       .        .        . 54 

Angles  of  a  Polygon 56 

Quadrilaterals 58 

Parallelograms 60 

Exercises  in  Invention 71 

3 


CONTENTS. 


BOOK  II. 
RATIO    AND    PROPORTION. 

PAGE 

Definitions 74 

Theorems 77 

BOOK  III. 

THE    CIRCLE. 

Definitions 85 

Chords,  Arcs,  etc 87 

Relative  Position  of  Circles 97 

Measurement  of  Angles 99 

Problems  in  Construction 107 

Exercises  in  Invention 124 

BOOK  IV. 
AREA    AND    RELATION    OF    POLYGONS. 

Definitions 126 

Areas 127 

Squares  on  Lines 133 

Projection 135 

Proportional  Lines 140 

Similarity  of  Polygons 143 

Relation  of  Polygons 160 

Problems  in  Construction 164 

Exercises  in  Invention 177 

BOOK  V. 
REGULAR    POLYGONS   AND   THE    CIRCLE. 

Definitions 179 

Relation    between    the    Circumference   and    the 

Diameter  of  a  Circle 183 

Problems  in  Construction 188 

Exercises  in  Invention 195 


PREFACE 


This  little  volume  has  been  prepared  with  a  view  to  furnish 
a  suitable  text-book  on  Plane  Geometry  for  Grammar  Schools, 
Preparatory  Schools,  etc. 

A  simple  method  of  designating  angles  has  been  adopted, 
and  recognized  symbols  have  been  freely  used  in  the  demon- 
strations, thus  bringing  the  several  steps  closely  together  and 
enabling  the  student  to  master  the  argument  with  ease.  The 
reasons  on  which  the  steps  of  an  argument  depend  are  not 
formally  given,  but  are  referred  to  by  numbers  indicating  the 
sections  in  which  they  are  found :  it  is  believed  that  the  pupil 
will  impress  the  principles  most  firmly  on  his  mind  by  fre- 
quency of  reference. 

No  valid  objection  can  be  offered  against  the  algebraic  form 
of  which  some  of  the  demonstrations  partake,  for  most  of  the 
axioms  laid  down  are  nothing  more  than  properties  of  the 
equation. 

No  apology  is  deemed  necessary  for  the  application  of  the 
Infinitesimal  method:  it  has  been  employed  whenever  it  gave 
directness,  brevity,  and  simplicity  to  the  demonstration. 

At  the  close  of  each  book,  except  the  second,  a  collection 
of  theorems  and  problems  has  been  placed  for  the  purpose  of 
giving  the  pupil  an  opportunity  to  exercise  his  originality  in 
demonstration  and  construction.  A  proper  use  of  these  exer- 
cises will  do  much  toward  stimulating  thought  and  awakening 
a  spirit  of  invention  in  the  pupil. 


6  PREFACE. 


During  the  preparation  of  this  treatise,  Diesterweg's  "Ele- 
mentare  Geometrie"  and  most  of  our  American  treatises  have 
been  freely  consulted. 

And  now,  this  little  work  is  respectfully  submitted  to  the  edu- 
cational public,  in  the  hope  that  it  may  at  least  merit  a  careful 
perusal. 

F.  IBACH. 


Philadelphia,  Pa.,  May,  1SS2. 


NOTE  TO  THE  TEACHER. 

In  recitation,  when  studying  a  book  for  the  first  time,  the  pupil 
should  be  required  to  draw  the  diagram  accurately  and  write  the 
demonstration  neatly  on  the  blackboard. 

Being  called  upon  to  recite,  he  enunciates  the  proposition  and 
gives  the  demonstration,  pointing  to  the  parts  of  the  diagram  as 
reference  is  made  to  them. 

In  review,  the  diagram  only  should  be  put  on  the  board. 


INTRODUCTION. 


SUBJECT-MATTER. 


The  accompanying  diagram  represents  a  block  of  granite, — a 
physical  solid,  of  regular  form. 

Such  a  block  has  six  flat  faces,  called  Surfaces.    It  has  also 
twelve  sharp  edges  in  which   these  surfaces 
meet,  called  Lines. 

It    has,   besides,    eight   sharp   corners  in 
which  these  lines  meet,  called  Points. 

If  the  block  be  removed,  we  can  imagine 
the  space  which  it  filled  to  have  the  same 
shape  and  size  as  the  block.  This  limited 
portion  of  space,  which  has  lefigth,  breadth,  and  thickness, 
is  called  a  Geometrical  Solid.  Its  boundaries  or  surfaces 
separate  it  from  surrounding  space,  and  have  length  and 
breadth  but  no  thickness.  The  boundaries  of  these  surfaces  are 
lines,  and  have  length  only.  The  limits  of  these  lines  are  points, 
and  have  position  only.  We  thus  come  in  three  steps  from 
solids  to  points,  which  have  no  magnitude.  Having  thus 
acquired  notions  of  solids,  surfaces,  lines,  and  points,  we  can 
easily  conceive  of  them  distinct  from  one  another.  It  is  of 
such  ideal  solids,  surfaces,  lines,  and  points  that  Geometry 
treats;  and  these  in  various  forms,  except  points,  are  called 
Geometrical  Magnitudes  or  Magnitudes  of  Space. 


ELEMENTS 

OF 

PLAICE  GlEOMETRT. 

BOOK^""" 


DEFIlSriTIO 

1.  Geometry  is  the  science  which  treats  of  the  properties 
and  relations  of  magnitudes  of  space. 

Space  has  extension  in  all  directions;  but  for  the  purpose  of 
determining  the  size  of  portions  of  space,  we  consider  it  as 
having  three  dimensions,  namely,  length,  breadth,  and  thickness. 

2.  A  Point  is  position  without  size. 

3.  A  ijine  is  that  which  has  but  one  dimension,  namely, 
length. 

A  line  may  be  conceived  as  traced  by  a  moving  point. 
Lines  are  straight  or  curved, 

4.  A  straight  Line  is  one  which 
has  the  same  direction  at  all  its 
points. 

5.  A  Ourvea  Line  is  one  which  changes  its  direction  at  all 
its  points.     When  the  sense  is  ob- 
vious, the  word  line,  alone,  is  used 
for   straight    line,   and    the   word 
curve,  alone,  for  curved  line. 

6.  A  Surface  is  that  which  has  only  two  dimensions,  length 
and  breadth. 

A  surface  may  be  conceived  as  generated  by  a  moving  line. 
Surfaces  are  plane  or  curved, 

9 


10       ELEMENTS  OF  PLANE  GEOMETRY. 

7.  A  Plane  Surface,  or  a  Platte, 

is  a  surface  ^vith  which  a  straight 
line  can  be  made  to  coincide  in 
any  direction. 

8.  A  Curved  Surface  is  a  surface  no  portion  of  which  is 
a  plane. 

9.  A  Solid  is  that  which  has  three  dimensions,  lengthy 
breadth,  and  thickness.  A  solid  may  be  conceived  as  generated 
by  a  moving  surface. 

Points,  lines,  surfaces,  and  solids  are  the  concepts  of 
Geometry,  and  may  be  said  to  constitute  the  subject-matter 
of  the  science. 

10.  A  Figure  is  some  definite  form  of  magnitude. 

11.  Lines,  surfaces,  and  solids  are  called  figures  when 
reference  is  had  to  their  form. 

12.  A  Plane  Figure  is  one,  all  of  whose  points  are  in  the 
same  plane. 

13.  Plane  Geometry  treats  of  plane  figures. 

14.  Equal  Figures  are  such  'as  have  the  same  form  and 
size,  that  is,  such  as  fill  exactly  the  same  space. 

15.  FQuivalent  Figures  are  such  as  have  equal  magnitudes. 

16.  Similar  Figures  are  such  as  have  the  same  form, 
although  they  may  have  difierent  magnitudes. 

17.  A  Plane  Angle,  or  an  Angle,  is  the  opening  between 
two  lines  which  meet  each  other.  The  point  in  which  the 
lines  meet  is  called  the  Vertex,  and  the  lines  are  called  the 
sides  of  the  angle.     A  plane  angle  is  a  species  of  surface. 

An  angle  is  designated  by  placing  a  letter  at  each  end  of 
its  sides,  and  one  at  its  vertex,  or  by  placing  a  small  letter  in 
it  near  the  vertex.  The  latter  is  the  method  employed  in 
this  book,  whenever  it  is  convenient.  In  reading,  when  there 
is  but  one  angle,  we  may  name  the  letter  at  the  vertex;  but 
when  there  are  two  or  more  vertices  at  the  same  point,  we 


ELEMENTS  OF  PLANE  GEOMETRY. 


11 


name  the  three  letters,  •with  the  one  at  the  vertex  between  the 
other  two:  we  may,  however,  in  either  case,  simply  name  the 
letter  placed  in  it.    Thus,  in  Fig.  1,  we  say  angle  C,  or  angle  a. 

Fig.  2. 


G 


-B 


D 


Q 


E 


In  Fig.  2,  G  being  tiie  common  vertex,  we  must  say  angle 
D  GF,  or  angle  b.  The  size  of  an  angle  depends  upon  the  exte^it 
of  opening  of  its  sides,  and  not  upon  the  length  of  the  sides. 

18.  Adjacent  Angles  are  SUch 
as  have  a  common  vertex  and 
one  common  side  between  them. 
Thus,  the  angles  a  and  b  are  ad- 
jacent angles. 

19.  A  Right  Angle  is  an  angle 
included  between  two  straight  lines 
which  meet  each  other  so  as   to 

make  the  adjacent  angles  equal.     

Thus,  if  the  angles  a  and  b  are 
equal,  each  is  a  right  angle. 

20.  Perpendicular   Lines    are    SUch    as 
make  right  angles  with  each  other. 

21.  An  Actite  Angle  is  one  which  is  less 
than  a  right  angle;  as  angle  a. 

22.  An  Obtuse  Angle  is   one 

which    is   greater   than  a  right 
angle;  as  angle  ABC. 

Acute  and  obtuse  angles  are 
called  oblique  angles. 


12 


ELEMENTS  OF  PLANE  GEOMETRY. 


23.  OhliQue  ijinea  are  lines  which  are  not  perpendicular 
to  each  other,  and  which  meet  if  sufficiently  produced. 


24.  Two  angles  are  Conijplements  of 
each  other  when  their  sum  is  equal  to  a 
right  angle.  Thus,  angle  a  is  the  com- 
plement of  angle  b,  and  angle  b  is  the 
complement  of  angle  a. 


25.  Two  angles  are  Supple- 
nients  of  each  other  when  their 
sum  is  equal  to  two  right  angles. 
Thus,  angle  a  is  the  supplement 
of  angle  ABC,  and  angle  ABC 
is  the  supplement  of  angle  a. 


26.  Vertical  Anyles  are  such  as 
have  a  common  vertex,  and  their 
sides  lying  in  opposite  directions. 
Thus,  angles  a  and  b  are  vertical ; 
also  angles  c  and  d. 


-JU- 
c/d 


.A 


27.  If  two  lines  are  cut  by  a  third  line,  eight  angles  are 
formed,  which  are  named  as  follows: 

Angles  a,  b,  c,  and  d  are  Exterior    

Angles.  Angles  e,  /,  g,  and  h  are 
Interior  Angles,  The  pairs  of  an- 
gles a  and  d,  b  and  c,  are  Alternate 
Exterior  Angles.  The  pairs  of  an- 
gles e  and  h,  f  and  g,  are  Alternate  Interior  Angles.  The 
pairs  of  angles  a  and  g,  b  and  h,  e  and  c,  /  and  d,  are  Cor- 
responding Angles, 

28.  Parallel  Straight  Lines  are 

such  as  lie  in  the  same  plane  and 

cannot   meet   hoAV  far  soever  they     '~^ 

are  produced  either  way.     They  have  the  same  direction. 


ELEMENTS  OF  PLANE  GEOMETRY.  13 

29.  A  Circle  is  a  plane  figure  bounded  by  a  curve,  all  the 
points  of  which  are  equally  distant  from  a  point  within, 
called  the  Centre. 

The  Circumference  of  a  circle  is  the 
curve  which  bounds  it. 

A  Radius  of  a  circle  is  a  line  extend- 
ing from  the  centre  to  any  point  in  the 
circumference. 

•    The  diagram  represents  a  circle  whose 

centre  is  0.     The  curve  ABCD  is  the  circumference,  and 

the  line  OA  is  a  radius. 


14       ELEMENTS  OF  PLANE  GEOMETRY. 

DEFINITIONS    OF   MATHEMATICAL 
TEEMS. 

30.  A  netnonstration,  or  Proof,  is  a  course  of  reasoning 
by  which  the  truth  of  a  statement  is  deduced. 

31.  An  Axiotn  is  a  statement  of  a  truth  which  is  self- 
evident. 

32.  A  Thcoretn  is  a  statement  of  a  truth  which  is  to  be 
demonstrated. 

33.  A  JProbietn  is  a  statement  of  something  to  be  done. 

34.  A  Postulate  is  a  problem  whose  solution  is  self-evident. 

35.  Axioms,  theorems,  and  problems   are  called  Propo- 
sitions. 

36.  A  Corollary  is  a  statement  of  a  truth  which  is  a  direct 
inference  fr(jm  a  proposition. 

37.  An  Hypothesis  is  a  supposition  made  in  a  proposition 
or  in  a  demonstration. 

38.  A  Scholium  is  a  comment  on  one  or  more  propositions. 

39.  AXIOMS. 

1.  Things  which  are  equal  to  the  same  thing  are  equal  to 
each  other. 

2.  If  equals  are  added  to  equals,  the  sums  are  equal. 

3.  If  equals  are  subtracted  from  equals,  the  remainders 
are  equal. 

4.  If  equals  are  added  to  unequals,  the  sums  are  unequal. 

5.  If  equals  are  subtracted  from  unequals,  the  remainders 
are  unequal. 

6.  If  equals  are  multiplied  by  equals,  the  products  are 
equal. 

7.  If  equals  are  divided  by  equals,  the  quotients  are  equal. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  the  sum  of  all  its  parts. 


ELEMENTS  OF  PLANE  GEOMETRY.  15 

10.  Only  one  straight  line  can  join  two  points. 

11.  A  straight  line  is  the  shortest  distance  from  one  point 
to  another. 

12.  All  right  angles  are  equal. 

40.   POSTULATES. 

1.  A  straight  line  can  be  drawn  joining  any  two  points. 

2.  A  straight  line  can  be  produced  to  any  length. 

3.  From  the  greater  of  two  straight  lines,  a  part  can  be 
cut  equal  to  the  less. 

4.  In  a  plane  a  circumference  of  a  circle  can  be  described, 
with  any  point  as  a  centre,  and  any  distance  as  a  radius. 

5.  Figures  can  be  freely  moved  in  space  without  change 
of  form  or  size. 

41.  DEMONSTRATION. 

A  Demonstration  is  a  logical  process,  the  premises  being 
definitions  and  self-evident  and  previously  established  truths. 

There  are  two  methods  of  demonstration,  called  the  JOirect 
MetHod  and  the  indirect  Method, 

The  nirect  Method  proves  a  truth  by  referring  to  defi- 
nitions and  self-evident  and  previously  deduced  truths,  and 
concludes  directly  with  the  proof  of  the  truth  in  question. 

The  Indirect  Method  proves  a  truth  by  showing  that  a 
supposition  of  its  falsity  leads  to  an  absurdity; — called  also 
redudio  ad  absurdum. 

Pupils  frequently  fall  into  errors  of  demonstration.  Notable 
among  these  errors  are  Begging  the  Question  and  Reason- 
ing in  a  Circle, 

Begging  the  Qt€estion  is  a  form  of  argument  in  which 
the  truth  to  be  proved  is  assumed  as  established. 

Reasoning  in  a  Circle  is  a  form  of  argument  in  which  a 
truth  is  employed  to  prove  another  truth  on  which  the  former 
depends  for  its  proof 


16  ELEMENTS  OF  PLANE  GEOMETRY. 

42.   EXPLANATION    OF  SYMBOLS   AND 
ABBREVIATIONS. 

=  denotes  equality. 


+ 

« 

increased  by. 

— 

ii 

diminished  by. 

X 

li 

multiplied  by. 

-f- 

(I 

divided  by. 

(( 

therefore. 

II 

it 

parallel. 

lis 

ii 

parallels. 

z. 

<( 

angle. 

As 

t( 

angles. 

L 

(C 

right  angle. 

Ls 

(I 

right  angles. 

-L 

(C 

perpendicular. 

J_s 

il 

perpendiculars. 

> 

(( 

is  greater  than. 

< 

(t 

is  less  than. 

A 

li 

triangle. 

As 

ii 

triangles. 

RA 

(I 

right-angled  triangle. 

RAs 

ii 

right-angled  triangles. 

O 

ii 

circle. 

Os 

ii 

circles. 

o 

ii 

parallelogram. 

Os 

ii 

dene 

parallelograms. 

Ax. 

>tes  axiom. 

Cor. 

(( 

corollary. 

Cons 

(( 

construction. 

Hyp. 

(( 

hypothesis. 

Q.E.D.  " 

quod  erat  demonstrandum  (which  was  to 

be  demonstrated). 

Q.E.F.  " 

quod  erat  faciendum  (which  was  to  be  done). 

ELEMENTS  OF  PLANE  GEOMETRY.  17 


PEEPET^DICULAR  Al^D    OBLIQUE 
STRAIGHT  LINES. 


THEOREM  I. 

43.  At  a  point  in  a  straight  line,  only  one  perpendicular 
can  be  erected  to  that  line. 

Let  O  be  a  point  in  the  line  AB. 

E  D 


A  a  Bs 

To  prove  that  only  one  JL  can  he  erected  to  AB  at  C. 

Draw  the  oblique  line  CD. 

Revolve  CD  about  C  so  as  to  increase  Z.  a  and  decrease 
Z.  ACD. 

It  is  evident  that  in    one  position  of  CD,  as  EC,  the 
adjacent  Z-s  are  equal. 

But  then  EC  is  JL  to  AB.  (20) 

And  there  can  be  only  one  position  of  the  line  in  which  the 
adjacent  Z_8  are  equal; 

.  • .  only  one  _L  can  be  erected  to  AB  at  the  point  C.    Q.  E.  D. 

2* 


18 


ELEMENTS  OF  PLANE  GEOMETRY, 


THEOREM   II. 

44.   The  sum  of  the  two  adjacent  angles  formed  by  two  lines 
which  meet  equals  two  right  angles. 

Let  Z.S  a  and  ACD  be  formed  by  the   line  CD  meet- 
ing AB. 


0 


B 


To  prove  that  Z.  a  +  Z_  ^C'i)  =  2  Ls. 
Let  CE  be  _L  to  AB  at  C. 

L.a^  jL  ECD  =  a  L, 
and  /L  ACD—  Z.ECD  =  2iL. 

Add  ;  then       /La -^  Z.  ACD  =  2Lb.     (Ax.  2)  Q.  E.  D. 

45.  Cor.  1. — If  one  of  the  two  adjacent  angles  formed  by 
two  lines  which  meet  is  a  right  angle,  the  other  is  also  a  right 
angle. 

46.  Cor.  2. —  The  sum  of  all  the  angles  formed  at  a  common 
point  on  the  same  side  of  a  straight  line  equals  two  right 
angles. 


Thus,  Z.a4-Z.6+Z.c 
+  Z.  cZ  +  Z.e  =  2Ls. 


ELEMENTS  OF  PLANE  GEOMETRY.  19 


THEOREM   III. 

47.  Conversely. — If  the  sum  of  two  adjacent  angles  equals 
two  right  angles,  their  exterior  sidts  lie  in  the  same  straight 
line. 

Jjet  Z.  a -^  Z.  ACD  =  2Ls. 


E 

b/a 


0 


To  prove  that  A  C  and  B  C  lie  in  the  same  straight  line. 

Draw  EC. 

li  EG  and  BC  lie  in  the  same  straight  line, 

Z.  a  +  Z.  6  ==  2  Ls.  (44) 

But        Z-a-^  I.  AGD=2L9',  (Hyp.) 

Z.  a-\-Z-h  =  I.  a-\-l.ACD.  (Ax.  1) 

From  each  member  subtract  Z.  a. 

Then  Z.  6  =  Z_  ^  CD,  which  is  impossible. 

(Ax.  8) 
. * .    AC  and  B C  lie  in  the  same  straight  line.       Q.  E.  D. 


20       ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM   IV. 

48.  If  two  straight  lin6s  intersect,  the  opposite  or  vertical 
angles  are  equal. 

Let  AB  and  CD  intersect. 

C 


To  prove  that  Z-  a=^  /-  h. 

Z.  a  4-  ^  c  =  2  Ls,  (44) 

and               Z.  5  +  A  c  =  2  Ls;  (44) 

/^a-]rl-c=l.h^A.c.  (Ax.  1) 
From  each  member  subtract  Z_  c. 

Then  /L  a  =  /L  h. 

Likewise  we  can  prove  that  L.  c  =  I-  d.  Q.  E.  D. 

49.  Cor.  1. — Ij  two  straight  lines  intersect,  the  sum  of  the 
four  angles  formed  equals  four  right  angles. 

50.  Cor.  2. — The  sum  of  all  the  angles  that  can  be  formed 
at  a  common  point  equals  four  right  angles. 


ELEMENTS  OF  PLANE  GEOMETRY, 


21 


THEOREM  V. 

51.  From  a  point  without  a  straight  line,  only  one  perpendic- 
ular can  he  drawn  to  that  line.  . 


Let  P  be  a  point  without  AB. 
P 


D 


C 


To  prove  that  only  one  _L  can  he  drawn  from  Pto  AB. 

Draw  the  oblique  line  PC.  With  the  point  P  fixed, 
revolve  PC  so  as  to  decrease  Z.  a  and  increase  Z_  h,  while 
the  common  vertex  moves  in  the  direction  CA. 

At  some  position  of  the  line,  as  PD,  the  adjacent  angles 
are  equal. 

Then  PD  is  JL  to  AB.  (20) 

There  is  only  one  position  of  the  line  in  which  the  angles 
are  equal. 

only  one  _J_  can  be  drawn  from  P  to  AB.  Q.  E.  D. 


22 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  VI. 

52.  From  a  point  without  a  straight  line,  a  perpendicular  is 
the  shortest  distance  to  that  line. 

Let  AB  be  a  straight  line,  C  any  point  without  it,  CE  a 
J_,  and  CF  any  oblique  line. 


F\  E 


-B 


D 


To  prove  that  C E  <  CF. 


Produce  CE  to  D,  making  ED  =  CE,  and  draw  FD. 

On  AB  as  an  axis,  revolve  the  plane  of  CEF  till  it  falls 
in  the  plane  of  DEF. 

Since  Z.  s  a  and  h  are  Ls,  the  line  CE  takes  the  direction 
ED,  the  point  C  falling  on  D. 


or 


CE=ED; 
CE-ir  ED  =  2  CE. 

CF=FD; 
CFA-  FD  =  2  CF. 
CE+  ED<  CF-]-  FD, 
2  CE<:2  CF. 


Divide  by  2 ; 
then 


CE  <  CF. 


(Cons.) 
(Ax.  10) 

(Ax.  11) 
Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


23 


THEOREM    VII. 

63.  Any  point  in  the  perpendicular  erected  at  the  middle 
point  of  a  straight  line  is  equally  distant  from  the  extremities 
of  that  line. 

Let  P  be  any  point  in  CD  which  is  JL  to  AB  at  its  middle 
point  D,  and  let  ^P  and  BP  be  drawn. 


To  prove  that  AP  =  BP. 

On  CD  as  an  axis,  revolve  APD  till  it  falls  in  the  plane 
of  BPD. 


Since  Z_s  a  and  h  are  Ls,  and  AD 
AP  =  BP. 


BD,  A  falls  on  B. 
(Ax.  10)     Q.  E.  D. 


54.  Cor.  1. — If  a  point  is  equally  distant  from  the  extremi- 
ties of  a  straight  line,  it  lies  in  the  perpendicular  erected  at 
the  middle  point  of  that  line. 

55.  Cor.  2. — If  each  of  two  points  in  a  straight  line  is 
equally  distant  from  the  extremities  of  another  straight  line,  the 
former  is  perpendicular  to  the  latter  at  its  middle  point. 


24 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  VIII. 

56.  Any  point  without  the  perpendicular  erected  at  the 
middle  point  of  a  straight  line  is  unequally  distant  from  the 
extremities  of  that  line. 


Let  P  be  any  point  without 
the  line  CD  which  is  J_  to 
AB  at  its  middle  point.  Draw 
AP  cutting  CD  at  0,  and 
draw^O  and  ^P. 


To  prove  that  AP  >  PP. 

BO  +  OP>BP,  (Ax.  11) 

and  AO  =  BO.  (53) 

Substitute  A  0  for  its  equal  B  0. 
Then  A0-\-  OP  >  BP. 

But  AO-{^OP=AP; 

AP>BP.  Q.E.D. 


THEOREM    IX. 

57.  Two  oblique  lines  drawn  from  a  point  in  a  perpendicu- 
lar are  equal  if  they  cut  off  equal  distances  from  the  foot  of  the 
perpendicular. 

C 

Let  CD  be  J_  to  AB,  and 
CE  and  (7P oblique  lines  cut- 
ting off  ED  =  DF. 

To  prove  that  CE  =  CF.       ^ 

On  CD  as  an  axis,  revolve  CDE  till  it  falls  in  the  plane 
of  CDF. 

Since  /.g  a  and  b  are  Ls,  and  ED  =  DF,   E  falls  on  F. 
CE=CF.  (Ax.  10)     Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY.  25 


THEOREM    X. 

58.  The  sum  of  two  lines  drawn  from  a  point  to  the  extremi- 
ties of  a  straight  line  is  less  than  the  sum  of  two  other  lines 
similarly  drawn  and  enveloping  them. 

Let  AP  and  BP  be  two  lines  drawn  from  P  to  the  extremi- 
ties oi  AB,  and  let  AC  and  BChe  two  lines  drawn  similarly 
and  enveloping  AP  and  BP. 


To  prove  that  AP  ^  BP  <  AC  +  BC. 

Produce  AP  to  D,  a  point  in  BC. 

AP  ■\- PD  <  AC -\-  CD, 
and  BP<PD-\-  DB.  (Ax.  11) 

Add  the  inequalities. 
Then     AP-\-BP-^PD  <AC^  CD  ^  DB-^PD. 

Substitute  BC  for  its  equal  CD  +  DB,  and  subtract  PD 
from  each  member. 

Then  AP-\-BP<AC^Ba  Q.  E.  D. 


26 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    XI. 

59.  Of  two  oblique  lines  drawn  from  the  same  point,  that  is 
the  greater  which  terminates  at  the  greater  distance  from  the 
foot  of  the  perpendicular. 

Let  CO  be  J_  to  AB,  and  CE  and  CD  oblique  lines 
drawn  so  that  EO  >  DO. 


E\I)\ 


To  prove  that  CE  >  CD. 

Produce  CO  to  F,  making  OF  =  CO,  and  draw  EF 
andDF. 

Then,  as  in  (52),     CD  =  DF,  and  CE  =  EF. 

But  CE-\-EF>  CD  +  DF, 

or  2  C^  >  2  CD.  (58) 

Divide  each  member  by  2. 

Then  CE  >  CD.  Q.  E.  D. 

60.  Cor.  1. —  Two  equal  oblique  lines  terminate  at  equal 
distances  from  the  foot  of  the  perpendicular. 

61.  Cor.  2. —  Only  two  equal  straight  lines  can  be  drawn 
from  a  point  to  a  line;  and  of  two  unequal  oblique  lines,  the 
greater  terminates  at  the  greater  distance  from  the  foot  of 
the  perpendicular. 


ELEMENTS  OF  PLANE  GEOMETRY.  27 


PAEALLEL    STKAIGHT    LIInTES. 


THEOREM    XII. 

62.  If  two  parallel  lines  are  cut  by  a  third  line — 

I.   The  corresponding  angles  are  equal; 

II.   Tlie  alternate  interior  angles  are  equal; 

III.   The  sum  of  the  interior  angles  on  the  same  side  of  the 
secant  equals  two  right  angles. 

Let  the  lis  AB  and  CD  be  cut  by  the  line  EF. 

E 

A 0/a  T> 

A yt" B 

F 

I.  To  prove  that  Z_  a  =  Z.  &. 

Since  OB  and  PD  are  II,  they  have  the  same  direction  and 
open  equally  from  the  line  EF; 

Z.a=^b. 

Likewise  we  can  prove  that  /L  c  =  /-  d. 

II.  To  prove  that  /-  c  =  Z^  b. 

^c  =  Z.a.  (48) 

But                          A.  a  =  Z.b;  (Case  I.) 

/.c  =  Z-b.  (Ax.  1) 
Likewise  we  can  prove  that  /^  e  =  Z.  f. 


28  ELEMENTS  OF  PLANE  GEOMETRY. 

III.   To  prove  that  Z.  b  -\- Z.  e  =  2Ls. 

A  a  +  Z.  e  =  2  Ls.  (44) 

Substitute  Z-  b  for  its  equal  Z_  a. 

Then  Z.6 +Z- e  =  2L.. 
Likewise  we  can  prove  that  Z.  c  -j- /^  f  =  2Ls.    Q.  E.  D. 

63.  Cor. — If  a  straight  line  lies  in  the  same  plane  with  two 
parallels,  and  is  perpendicular  to  one  of  them,  it  is  perpendicu- 
lar to  the  other  also. 

THEOREM    XIII. 

64.  If  two  straight  lines  are  cut  by  a  third  line,  these  two 
lines  are  parallel — 

I.  If  the  corresponding  angles  are  equal; 

II:  If  the  alternate  interior  angles  are  equal; 

III.  If  the  sum  of  the  two  interior  angles  on  the  same  side  of 
the  secant  equals  two  right  angles. 

Let  the  straight  lines  AB  and  CD  he  cut  by  EF. 

E 


F 

I.  To  prove  that  AB  and  CD  are  II  if  I-  a=^  A^b. 

If  /^a  =  Z-b,    OB  and  FD  open  equally  from  EF,  and 
hence  have  the  same  direction,  or  are  11.  (28) 

II.  To  prove  that  AB  and  CD  are  W  if  Z^  d  =  Z.  b. 

If  Z.a  =  Z.b,  AB  and  CD  are  II.  (Case  I.) 

But  Z.d  =  Z.  a;  (48) 

if  Z.d  =  Z.b,  AB  and  CD  are  II. 


ELEMENTS  OF  PLANE  GEOMETRY. 


29 


III.  To  prove  that  AB  and  CD  are  W  if  Z,  c  -^  Z.  b  =  2  Ls. 


l.a-{-  Z-c  =  2U; 
Z^c  -\-  Z.  b  =  2Ls; 
Z.a-{-Z.c=Z-G-{-Z-b. 

Subtract  Z_  c  from  each  member. 
Then  Z.  a  =  Z.  b. 

But  then  the  lines  are  II ; 
.     if     /.  c  -f  Z.  6  =  2  Ls,  ^^  and  CD  are  II 


(44) 
(Hyp.) 

(Ax.  1) 


(Case  I.) 
Q.E.D. 


65.  Cor. — If  two  lines  in  the  same  plane  are  perpendicular 
to  the  same  line,  they  are  parallel. 


THEOREM    XIV. 

66.  If  two  straight  lines  are  parallel  to  a  third  line,  they 
are  parallel  to  each  other. 


Let  AB  and  CD  be 


to  EF. 
G 


H 


To  prove  that  AB  is  \\  to  CD. 
Let  GH  be  JL  to  EF; 
Then  OH  is  JL  to  AB. 
GH  is  also  JL  to  CD; 
.    AB  is  !I  to  CD. 


(63) 

(63) 

(65)     Q.E.D. 


30  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XV. 

67.   Two  parallel  lines  are  everywhere  equally  distant  from 
each  other. 

Let^jB  and  CD  be  11,  and  GO  and  jffP-Ls  to  CD  drawn 
from  any  two  points  in  AB. 


To  prove  that  GO  =  HP. 

Let  the  _-L  EF  be  drawn  at  the  middle  point  between  G 
and  H. 

GO,  HP,  and  EF  are  _L  to  both  lis.  (63) 

On  EF  as  an  axis,  revolve  the  part  of  the  plane  on  the 
right  of  EF  till  it  falls  in  the  part  on  the  left  of  EF. 

Z.S  a  and  h  are  Ls,  and  EH  =  EG; 

/f  falls  on  G^. 

Z-  s  c  and  d  are  Ls ;  . 

.-.  HP  takes  the  direction  GO,  and  P  falls  in  G^O,  or  GO 
produced. 

Z.8  m  and  n  are  Ls; 
P  falls  in  the  line  FC 

Now  P  falling  in  both  GO  and  FC,  must  fall  at  their 
intersection  0; 

GO=HP.  Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


31 


EQUALITY   OF  AI^TGLES. 


THEOREM  XVI. 

68.  Two  angles  are  equal  if  their  sides  are  respectively 
parallel  and  lying  in  the  same  or  in  opposite  directions  from 
their  vertices. 

I.  Let  AB  and  BC,  the  sides  of  Z.  a,  be  respectively  II  to 
DE  and  EF,  the  sides  of  Z.  6. 

L 


To  prove  that  Z.  a  =  Z.  6. 

If  necessary,  produce  two  sides  not  II  till  they  intersect,  as 
at  G. 


Then 

/-  a  =  /-  c, 

and 

^b=  Z.c; 

(62) 

.-. 

L.a=  /_h. 

(Ax.  1) 

II.  Let  LM  and  ifiV,  the  sides  of  Z.  d,  be  respectively  11 
to  PQ  and  OP^  the  sides  of  Z_  e. 

To  prove  that  /-  d  =  /-  e. 

If  necessary,  produce  two  sides  not  II  till  they  intersect,  as 
ati?. 


Then 

^d=L,f, 

and 

Z.e=  A.f; 

(62) 

l.dr=  l_e. 

(Ax.  1)  Q.  E.  D. 

32  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XVII. 

69.  Two  angles  are  supplements  of  each  other  if  they  have 
two  sides  parallel  and  lying  in  the  same  direction,  and  the 
other  sides  parallel  and  lying  in  opposite  directions  from  their 
vertices. 

Let  Z.S  a  and  e  have  AC  and  EF  II  and  lying  in  the  same 
direction,  and  AB  and  ED  II  and  lying  in  opposite  directions 
from  their  vertices. 


To  prove  that  Z.  s  a  and  e  are  supplements  of  each  other. 

If  necessary,  produce  two  sides  not  II  till  they  intersect,  as 
at  G. 

Then  Z.  a  =  Z.  5,  and  Z.  ci  =  Z.  6.  (62) 

But       Z.  s  6  and  d  are  supplements  of  each  other ; 

Z.S  a  and  e  are  supplements  of  each  other. 

Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


33 


THEOREM  XVin. 

70.   T\vo  angles  having  their  sides  respectively  perpendicular 
to  each  other  are  either  equal  or  supplements  of  each  other. 

Let  the  Z.S  a  and  b  have  the  sides  BA  and  BC  respec- 
tively J-  to  EF^ud  ED. 

M  0  A 


\ 

/ 

\ 

\ 

X 

N    \ 

/ 

\     \ 

/ 

\    \c 

e,  / 

\^ 

Ya 

5      E 

\ 

'\^ 

h\ 

^\ 

\ 

d 

G 

\ 

I 

) 

> 

To  prove  that  Z.  s  a  and  b  are  either  equal  or  supplements  of 
each  other. 

At  B,  let  BO  and  BM  be  drawn  respectively  II  to  ED 
and  EF,  and  let  G^iV'  be  drawn  II  to  ^i^  or  BM. 

Then  I.  c  =  L.b,  (68) 

and  Z.  c  is  the  supplement  of  Z.  d.  (69) 

Now     Z-c-{-Z^e  =  ?i\^,  and  Z_e-|-  Z_a  =  aL; 

Z-c  -f-Z-e=Z_e+  /.a. 

Subtract  Z.  e  from  each  member. 

Then  /L  c  =  /^  a. 

But  Z.  c=  I-  b; 

Z.a=Z-b. 

Also,  since         Z.  c  is  the  supplement  of  Z_  d^ 

Z.  a  is  the  supplement  of  Z.  c?.  Q.  E.  D. 


34 


ELEMENTS  OF  PLANE  OEOMETBY 


TEIANGLES. 


DEFINITIONS. 

71.  A  Triangle  is  a  plane  figure  bounded  by  three  straight 
lines;  as  ABC. 

The  Sides  of  a  triangle  are  the  bounding 
lines. 


The  Angles  of  a  triangle  are  the  an| 
formed  by  the  sides  meeting  one  another.    A 
triangle  has  six  parts, — three  sides  and  three 
angles. 

The  nase  of  a  triangle  is  the  side  upon  which  it  is  sup- 
posed to  stand. 

The  Vertical  Angle  of  a  triangle  is  the  angle  opposite  the 
base.  An  Exterior  Angle  of  a  triangle  is  an  angle  formed 
by  a  side  and  an  adjacent  side 
produced ;  as  Z_  a. 

The  Vertex  of  a  triangle  is 
the  angular  point  at  the  ver- 
tical angle. 

The  Aititudti  of  a  triangle 
is  the  perpendicular  distance 

from  the  vertex  to  the  base,  or  to  the  base  produced.    Thus, 
CD  is  the  altitude  of  both  the  triangles 
ABC  and  EBC 

A  Medial  JLine  of  a  triangle  is  a  line 
drawn  from  a  vertex  to  the  middle  of  the 
opposite  side. 


72.  Triangles  are  classified 
sides  and  angles. 


to  their 


E  B 


ELEMENTS  OF  PLANE  GEOMETRY. 


35 


A  Scaietie  Triangle  is  one  which  has  no  two  sides  equal. 
An  Isosceles  Triangle  is  one  which  has  two  sides  equal. 


An  Equilateral  Triangle  is  one  which  has  all  its  sides 
equal. 

An  Acute- Angled  Triangle  is  one  which,  has  three  acute 
angles. 


A  uigHt- Angled  Triangle  is  one  which  has  one  right 
angle. 

An  Obtuse- Angled  Triangle  is  one  which  has  one  obtuse 
angle. 

An  Equiangular  Triangle  is  one  which  has  all  its  angles 
equal. 


36 


ELEMENTS  OF  PLANE  GEOMETRY. 


RELATION    BETWEEN    THE    SIDES 
OF  A  TRIANGLE. 


THEOREM   XIX. 

73.  Any  side  of  a  triangle  is  greater  than   the  difference 
between  the  other  two  sides. 

Let  ABC  be  any  A. 


To  prove  that  AC>  AB  —  BC. 

AB<AC-\-Ba 
Subtract  BC  from  each  member. 
Then  AB  —  BC<AC, 


or 


AOAB—BC. 


(Ax.  11) 


Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


37 


THEOREM   XX. 

74.  The  sum  of  the  three  lines  drawn  from  a  point  tuithin  a 
triangle  to  the  vertices  is  greater  than  half  the  sum  of  the  sides 
of  the  triangle. 

Let  m,  n,  o,  be  the  three  lines  drawn  from  any  point  P 
in  the  ^  ABC,  and  a,  b,  c,  the  sides  of  the  A. 


To  prove  that  m  -^  n  -\-  o  >  ^ 

m  +  0  >  a, 

n  -\-  0  ^  c, 

and  m  -\-  n'>  b. 

Add  these  inequalities. 

Then        2m -^  2n -\-  2o  >  a -{-  b  -\-  c, 

a  -\-  b  -\-  c 


(Ax.  11) 


or 


m  -|-  n  +  0  > 


Q.E.D. 


38 


ELEMENTS  OF  PLANE  GEOMETRY. 


MEDIAL    LINES. 


THEOREM    XXI. 

75.  The  sum  of  the  three  medial  lines  of  a  triangle  is  greater 
than  half  the  sum  of  the  sides  of  the  triangle.         q 


In  the  Z\  ABC,  let  a,  b,  c,  be 

/ 

he  sides,  and  m,  n,  o,  the  medial 
ines. 

A 

A 

/             0^ 

.  To  prove  that  m  -\-  n  -{-  o  ^           ^ 

m>  a  — 2 

m>6-2 

^          h 

> 

<'>*-L 

Add  these  inequalities. 

Then        2m  +  2n  +  2o  >  a  +  6  +  c, 

a  +  6  -f-  c 


(73) 


or 


m  +  n  -f-  0  > 


Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY.  39 


AI^GLES    OF  A  TRIANGLE. 


THEOREM  XXII. 

76.  The  sum  of  the  three  angles  of  a  triangle  equals  two  right 
angles. 

0  D 


Let  ^^C  be  any  A. 


B  F 

To  prove  that  Z^  a  -{-  Z-  h  -\-  Z^  c  ^=  1  \_s. 

Produce  AB,  and  let  DBhQ  I!  to  AC. 
Then        A  m  +  Z.  w  +  A  a  =  2  Ls.      ^  (46) 

But  jL  m=^  Z-h, 

and  Z-  n  =  Z^  c.  (62) 

For  Z.  8  m  and  n,  substitute  their  equals  Z.  s  6  and  c. 
Then  Z.  a  +  Z.  6  -I-  ^L  c  -=  2  Ls.  Q.  E.  D. 

77.  Cor.  1. —  The  sum  of  two  angles  of  a  triangle  being 
given,  the  third  can  he  found  by  subtracting  their  sum  from 
two  right  angles. 

78.  Cor.  2. — If  two  angles  of  a  triangle  are  respectively 
equal  to  two  angles  of  another y  the  third  angles  are  also  equal. 

79.  Cor.  3. — In  any  triangle,  there  can  be  but  one  right 
angle,  or  but  one  obtuse  angle. 

80.  Cor.  4. — In  any  right-angled  triangle,  the  sum  of  the 
acute  angles  equals  a  right  angle;  that  is,  they  are  complements 
of  each  other. 

81.  Cor.  5. — In  an  equiangular  triangle,  each  angle  equals 
one-third  of  two  right  angles. 


40  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XXIII. 

82.  An  exterior  angle  of  a  triangle  equals  the  sum  of  the 
two  interior  non-adjacent  angles. 

Let  ABC  be  any  A,  and  a  an  exterior  Z.. 

C 


A  B 

To  prove  that  Z^a=Z^c-\-/-d. 

Z.  a  +  ^  6  =  2  Ls,    .  (44) 

and        Z.  6  +  ^  c  +  Z.  cZ  =:  2  Ls;  (7.6) 

l.a-^/Lh=Z-h^A.c-^Z.d. 
Subtract  Z_  h  from  each  member ; 
then  Z.  a  =  Z.  c  +  Z.  d  Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


41 


EQUALITY  OF  TEIANGLES. 


THEOREM  XXIV. 

83.  Two  triangles  are  equal  in  all  their  parts  if  two  sides  and 
the  included  angle  of  the  one  are  respectively  equal  to  two  sides 
and  the  included  angle  of  the  other. 

In  the  As  ABC  and  DEF,  let  AB  =DE,  AC  =  DF, 
and  /-  a  =  /-  b    . 


A  B  r> 

To  prove  that  A  ABC  =  /^  DEF. 

Place  the  A  ABC  upon  DEF,  applying  Z.  a  to  its  equal 
Z.  h,  AB  on  its  equal  DE,  and  AC oti  its  equal  DF. 

The  points  C  and  B  fall  on  the  points  F  and  E; 

BC=EF.  (Ax.  10) 

and  ^     A  ^^C=  A  DEF.  (14) 

Q.E.D. 


42 


ELEMENTS  OF  PLANE  GEOMETRY, 


THEOREM  XXV. 

84.  Two  triangles  are  equal  in  all  their  parts  if  two  angles 
and  the  included  side  of  the  one  are  respectively  equal  to  two 
angles  and  the  included  side  of  the  other. 

In  the  As  ABC  and  DEF,  let  Z.  a  =  Z.  c?,  Z.6  = 
Z.  e,  and  AB  =  DE. 


To  prove  that  A  ABC=  Z^  DEF. 

Place  the  A  ABC  upon  the  A  DEF,  applying  AB  to  its 
equal  DE,  the  point  A  on  D,  and  the  point  B  on  E. 

Since   Z.  a  =  /-  d,  AC  takes  the  direction  DF,  and  O 
falls  somewhere  in  DF  or  DF  produced. 

Since  Z.  b  =  Z.  e,  BC  takes  the  direction  EF,  and   C 
falls  in  EE'  or  EF  produced; 

.  • .  the  point  C,  falling  in  both  DF  and  EF,  falls  at  their 
intersection  F; 

A  ABC  =  A  DEF.  (14)     Q.  E.  D. 


85.  Cor. — If  a  triangle  has  a  side,  its  opposite  angle,  and 
one  adjacent  angle,  respectively  equal  to  the  corresponding  parts 
of  another  triangle,  the  triangles  are  equal. 


ELEMENTS  OF  PLANE  GEOMETRY.  43 


THEOREM  XXVI. 

86.  Two  triangles  are  equal  in  all  their  parts  if  the  three 
sides  of  the  one  are  respectively  equal  to  the  three  sides  of  the 
other. 

In  the  As  ABC  and  DEF,  let  AC  =  DF,  BC  =  EF, 
and  AB  =  DE. 


To  prove  that  A  ABC  =  ^  DEF. 

Place  ABC  in  the  position  DEG,  AB  in  its  equal  DE, 
and  the  Z.s  a  and  c  adjacent  to  the  Z_  s  6  and  d. 

Draw  FO  cutting  DE  at  P. 

DG  =  DF,  and  EG  =  EF;  (Hyp.) 

the  points  D  and  E  are  equally  distant  from  F  and  G, 
and  DE  is  JL  to  EG  at  its  middle  point.  (55) 

On  DE  as  an  axis,  revolve  DEG  till  it  falls  in  the  plane 
of  DEF. 

Then  the  point  G  falls  on  jP,  since  PG  =  PF; 

A  ABC  =  A  DEF.  (14)       (Q.  E.  D.) 

87.  Cor. — In  equal  triangles,  the  equal  angles  lie  opposite  the 
equal  sides. 

88.  Scholium. — The  statement,  two  triangles  are  equal, 
means  that  the  six  parts  of  the  one  are  respectively  equal  to 
the  six  parts  of  the  other. 


44 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XXVII. 

89.  If  two  triangles  have  two  sides  of  the  one  respectively 
equal  to  two  sides  of  the  other,  and  the  included  angles  unequal, 
the  third  sides  are  unequal,  ayid  the  greater  third  side  is  in  the 
triangle  having  the  greater  included  angle. 

In  the  As  ABC  and  DEF,  let  ^C  =  DF,  CB  =  FE, 

and  ACB  >  Z.  c. 


To  prove  that  AB  >  DE. 

Place  the  A  DEF  so  that  EF  falls  in  its  equal  BC. 
Let  CH  bisect  Z.  ECA,  and  draw  EH. 

CE=CA,  (Hyp.) 

CH=-  CH, 
and  /-  a  =  A.  b;  (Cons.) 

ACHE=  A  CHA,  (83) 

and  EH=AH. 

Now  HB  +  EH>  EB. 

Substitute  AH  and  DE  for  their  equals  EH  and  EB. 
Then  AH-^  HB>  DE, 


or 


AB  >  DE. 


Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


45 


THEOREM  XXVin. 

90.  Conversely. — If  two  sides  of  a  triangle  are  respectively 
equal  to  two  sides  of  another,  and  the  third  sides  unequal,  the 
angle  opposite  the  third  side  is  greater  in  the  triangle  having  the 
greater  third  side. 

In  the  As  ABC  and  DEF,  let  AC=DF,  BC  =  EF, 
and  AB  >  DE. 


To  prove  that  /L  a'>  Z.  6. 

If         Z.  a  =  21  6,  A  ABC  =  A  DEF,  (83) 

and  AB  =  DE. 

If  /.a<  Z-  b,AB<  DE.  (89) 

But  both  these  conclusions,  being  contrary  to  the  hypoth- 
esis, are  absurd. 

.  • .   /-a  cannot  equal  Z-  b,  and  cannot  be  less  than  Z.  b, 
Z.a>  ^b.  Q.E.D. 


46 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM   XXIX. 

91.  Two  right-angled  triangles  are  equal  in  all  their  parts  if 
the  hypothenuse  and  one  side  of  the  one  are  respec^vely  equal  to 
the  hypothenuse  and  one  side  of  the  other. 


In  the  RAs  ABC  and  DEF,  let  the  hypothenuse  BC  = 
EF.  and  AC  =DF. 


D 


E 


To  prove  that  R/\  ABC  =  BA  DBF. 

Place  ABC  on  DBF  so  that  ^C  falls  in  its  equal  DF. 

Z-B  a  and  b  are  Ls;  (Hyp.) 

AB  takes  the  direction  DE, 
and  B  falls  on  E;  (60) 

R  A  ABC  =  R A  DBF.  (14)     Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


47 


THEORE 


92.  Two  right-angled  triangles  are  equal  in  all  their  paris  if 
the  hypothenuse  and  one  acute  angle  of  the  one  are  respectively 
equal  to  the  hypothenuse  and  one  acute  angle  of  the  other. 

In  the  RAs  ABC  and  DEF,  let  BC  =  EF,  and  ^  a  = 
Z.  b. 


To  prove  that  i?A  ABC=EZ^  DEF. 


and 


BC=EF, 

(Hyp.) 

/_a=  l.h, 

(Hyp.) 

Z.c=  Ad; 

(78) 

RA  ABC=B,/\  DEF. 

(84)    Q.E.D. 

93.  Cor. — Two  right-angled  triangles  are  equal  if  a  side 
and  an  acute  angle  of  the  07ie  are  respectively  equal  to  a  side 
and  an  acute  angle  of  the  other. 


48       ELEMENTS  OF  PLANE  GEOMETBT. 


RELATION  BETWEEN  THE  PAETS 
OF  A  TRIANGLE. 


THEOREM  XXXI. 

J- 

94.  In  an  isosceles  triangle,  the  angles  opposite  the  equal  sides 
are  equal. 

In  the   isosceles  A  ABC,  lei  AC  and  BC   he  the  equal 
sides. 

C 


A  D  B 

To  prove  that  L.  a  =^  L.  h. 

Let  CD  be  drawn  to  the  middle  of  AB. 

AC=BC,  (Hyp.) 

AD  =  BD,  (Cons.) 
and                         CD  =  CD; 

A  ADC=  A  BDC,  (86) 

and                         /.  a=  Z.  b.  Q.  E.  D. 

95.  Cor.  1. —  The  straight  line  joining  the  vertex  and  the 
middle  of  the  base  of  an  isosceles  triangle  bisects  the  vertical 
angle  and  is  perpendicular  to  the  base. 

96.  Cor.  2. —  The  straight  line  which  bisects  the  vertical  angle 
of  an  isosceles  triangle  bisects  the  base  at  right  angles. 

97.  Cor.  3. — Any  equilateral  triangle  is  equiangular. 


ELEMENTS  OF  PLANE  GEOMETRY. 


49 


THEOREM   XXXII. 

98.  Conversely. — IJ  two  angles  of  a  triangle  are  equal,  the 
sides  opposite  them  are  equal,  and  the  triangle  is  isosceles. 


In  the  A  ABC,  let  Z.  a 


To  prove  that  AC  =  BC. 

Let  CD  be  J_  to  AB. 

/.a:=  Z.b,  (Hyp.) 

and  CD  =  CD; 

B.AADC=B.ABDC,  (93) 

and  AC=BC.  Q.  E.  D. 


THEOREM   XXXIII. 

99.   Of  two  angles  of  a  triangle,  the  greater  is  opposite  the 
greater  side.  O. 


In  the  A  ABC,  let 
CB  >  AB. 


To  prove  that  Z.  BAO  L.  e. 
Cut  off  BD  =  AB,  and  draw  AD. 

Z.  a=  I-  d.  (94) 

But  Z.  d>  Z.  c;      >  (82) 

Z.  a>  ZL  c. 
And  much  more  is  Z.  BAO  Z.  c.  Q.  E.  D. 


50 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XXXIV. 

100.  Conversely. —  Of  two  sides  of  a  triangle,  the  greater 
is  opposite  the  greater  angle. 

In  the  A  ABC,  let  /L  BAO  Z.  c. 


To  prove  that  BO  AB. 

Let  AD  be  drawn  so  as  to  make  Z-a^=  JL  c. 

Then  AD  ==  CD. 

Now  AD  -^  BD>  AB. 

Substitute  CD  for  its  equal  AD. 

Then  CD  -\^  BD>  AB, 

or  BO  AB. 


(98) 
(Ax.  11) 


Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


51 


BISECTOES   OF  ANGLES. 


THEOREM    XXXV. 

101.  Any  point  in  the  bisector  of  a?i  angle  is  equally  distant 
from  the  sides  of  the  angle. 

Let  BF  be  the  bisector  of  the  Z.  ABC,  P  any  point  in  it, 
and  PD  and  PE  J_s  to  AB  and  BC. 


To  prove  that  PD  =  PE. 


Z_3  a  and  h  are  Ls, 

/-c=  jLd, 

(Hyp.) 

and 

BP  =  BP; 

R  A  BDP  =  EA  BEP, 

(92) 

and 

PD  =  PE. 

Q.  E.  D. 

102.  Cor. — Any  point  in  an  angle  equally  distant  from  the 
sides  lies  in  the  bisector  of  the  angle. 


62  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM   XXXVI. 

103.   The  bisectors  of  the  angles  of  a  triangle  meet  in  a 
common  point. 

Let  AEy  BFj  and  CD  be  the  bisectors  of  the  Z.s  of  the 
A  ABC. 


A  B 

To  prove  that  AE,  BF,  and  CD  meet  in  a  common  point. 

Let  AE  and  BF  meet  in  a  point,  as  0. 
Then   0  is  equally  distant  from  AB  and  AC;  also  from 
ABandBC;  (101) 

.*.  0  is  equally  distant  from  J. C  and  BC,  and  lies  in  CD; 

(102) 

.*.  the  bisectors  AE,  BF,  and  CD  meet  in  a  common  point. 

Q.  E.  B. 

104.  Cor. — The  point  in  which  the  bisectors  of  the  angles 
of  a  triangle  meet  is  equally  distant  from  the  three  sides  of  the 
triangle. 


ELEMENTS  OF  PLANE  OEOMETBY.  63 


THEOREM   XXXVII. 

105.   The  perpendiculars  erected  at  the  middle  points  of  the 
sides  of  a  triangle  meet  in  a  common  point. 

In  the  A  ABC,  let  DH,  FG,  and  EM  be  respectively  _L 
to  AC,  AB,aiid  BC,  at  their  middle  points. 


To  prove  that  DH,  FG,  and  EM  meet  at  a  common  point. 

The  J_s  DH  and  FG  meet  in  some  point,  as  0,  other- 
wise they  would  be  II,  and  AC  and  AB,  the  JLs  to  these  lis, 
would  lie  in  the  same  straight  line,  which  is  impossible. 

Now  0  is  equally   distant   from  A   and  C;    also   from   A 
and  B;  (53) 

.-.   0  is  equally  distant  from  C  and  B,  and  must  lie  in 
EM  (54).     That  is,  the  J_  EM  passes  through  0; 

.  • .  DH,  FG,  and  EM  meet  in  a  common  point.        Q.  E.  D. 


106.  CoE. — The  common  point  of  the  perpendiculars  erected 
at  the  middle  points  of  the  sides  of  a  triangle  is  equally  distant 
from  the  vertices  of  the  triangle. 

5* 


54      ELEMENTS  OF  PLANE  GEOMETRY. 


POLYGONS. 


DEFINITIONS. 

107.  A  roiygon  is  a  plane   figure   bounded  by  straight 
lines.     The  bounding  lines  are  the  sides  of  the  polygon. 

The  Perimeter  of  a  polygon  is  the  sum 
of  the  bounding  lines.  The  angles  which 
the  adjacent  sides  make  with  each  other 
are  the  angles  of  the  polygon. 

A  niagonai  of  a  polygon  is  a  line  join- 
ing two  non-adjacent  angles. 

Note. — Let  the  pupil  illustrate. 

108.  An  Equilateral  Polygon,  is  one  all  of  whose  sides 
are  equal. 

109.  An  EquianguUir  Polygon  is  one  all  of  whose  angles 
are  equal. 

Two  polygons  are  mutually  equilateral  when  their  sides  are 
respectively  equal. 

Two  polygons  are  mutually  equiangular  when  their  angles 
are  respectively  equal. 

Homologous  sides  or  angles  are  those  which  are  similarly 

placed. 

110.  A  Convex  Polygon  is  one  no   side   of  which  when 
produced  can  enter  the  surface  bounded  by  the  perimeter. 

Each  angle  of  such  a  polygon  is  called  a  salient  angle. 


ELEMENTS  OF  PLANE  GEOMETRY. 


55 


111.  A  Concave  Polygon  is  one  of  which  two  or  more 
sides,  when  produced,  will  enter  the  space  enclosed  by  the 
perimeter. 


1         Convex        1 
1       Polygon.       1 


The  angle  ^  OC  is  called  a  re-entrant  angle. 

112.  By  drawing  diagonals  from  the  vertex  of  any  angle 
of  a  polygon,  it  may  be  divided  into  as  many  triangles  as  it 
has  sides  less  two. 


113.  A  polygon  of  three  sides  is  a  Triangle;  of  four,  a 
Quadrilateral;  of  five,  a  Pentagon;  of  six,  a  Hexagon;  of 
seven,  a  Heptagon ;  of  eight,  an  Octagon ;  of  nine,  a  Nona- 
gon ;  of  ten,  a  Decagon ;  of  twelve,  a  Dodecagon. 


66       ELEMENTS  OF  PLANE  GEOMETRY. 


ANGLES  OF  A  POLYGOI^. 


THEOREM    XXXVIII. 

114.  The  sum  of  all  the  angles  of  any  polygon  equals  two 
right  angles  taken  as  many  times  less  two  as  the  polygon  has 


Let  ABCDEF  be  a  polygon  of  w  sides. 

E 


^D 


To  prove  that  Z.  AFE  +  Z.  FED  -}-  Z.  EDO,  etc.  = 

2  Ls  (n— 2) 
From  any  vertex,  as  F,  draw  the  diagonals  FB,  FC,  and 
FD. 

Then  we  have  (ti— 2)  As.  (112) 

The  sum  of  the  Z.s  of  the  As  =  the  sum  of  the  Z.sof 
the  polygon. 

But  the  sum  of  the  Z.s  of  a  A  =  2  Ls;  (76) 

the  sum  of  the  Z_s  of  the  polygon  =  2  Ls  (n  —  2). 

Q.E.D. 
115.  Cor. — The  sum  of  the  angles  of  a  quadrilateral  is  4  Ls; 
of  a  pentagon,  6  Ls ;  of  a  hexagon,  8  Ls,  etc. 


ELEMENTS  OF  PLANE  GEOMETRY.  57 


THEOREM  XXXIX. 

116.  If  the  sides  of  a  convex  polygon  are  produced  so  as  to 
form  one  exterior  angle  at  each  vertex,  the  sum  of  the  exterior 
angles  equals  four  right  angles. 

Let  ABODE  be  a  polygon  oin  sides,  and  let  the  sides  be 
produced  so  as  to  form  the  exterior  Z_  s  a,  6,  c,  cZ,  e. 


a         B 


To  prove  that  /.a+Z.5  +  Z.c+^c^+/.e  =  4Ls. 

At  each  vertex  there  are  two  Z.s  whose  sum  =  2  Ls;  (44) 

and  since  there  are  as  many  vertices  as  there  are  sides, 
we  have  n  X  2  Ls. 

But  the  sum  of  the  interior  Z_s  ;=  2  Ls  (w  —  2);        (114) 

./:^a^l.h-{-/Lc-\-/-d-^A.e=n  X  2  Ls  — 

2  Ls  (n  —  2) 

=  2nLs  — 2nL8  +  4Ls. 

=  4Ls. 

Q.E.D. 


58       ELEMENTS  OF  PLANE  GEOMETRY. 


QUADEILATERALS. 


IDEFINITIOlSrS. 

117.  A  Quadrilateral  is  a  polygon  of  four  sides. 

118.  There   are   three   classes   of   quadrilaterals,   namely, 
Trapeziums,  Trapeisoids,  and  Parallelograms. 

119.  A  Trapezium  is  a  quadrilateral  which  has  no  two 
of  its  sides  parallel. 

120.  A  Trapemoia  is  a  quadrilateral  which  has  two  of  its 
sides  parallel. 

The  parallel  sides  are  called  the  bases. 

121.  A  Parallelogram  is  a  quadrilateral  which   has   its 
opposite  sides  parallel. 

The  side  upon  which  a  parallelogram  is  supposed  to  stand 
and  the  opposite  side  are  called  the  bases. 

122.  A  Rectangle  is  a  parallelogram  whose  angles  are 
right  angles. 

123.  A  SQuare  is  an  equilateral  rectangle. 

124.  A  Ritomhoia  is  a  parallelogram  whose  angles  are 
oblique. 


ELEMENTS  OF  PLANE  GEOMETRY. 


59 


125.  A  Mhombuft  is  an  equilateral  rhomboid. 


z 

/    Trapeziom 

. 

/             Trapezoid.            \ 

\                  ParaUelogram.               \ 

Rectangle. 

\                                 \ 

Square. 

\          Rhomboid.        \ 

I 

126.  A  Diagonal  of  a  parallelogram  is  a  line  joining  any 
two  opposite  vertices. 


Note. — Let  the  pupil  illustrate. 

127.     The  Aitituae  ;of  a  parallelogram  or  trapezoid  is  the 
perpendicular  distance  between  its  bases. 


60  ELEMENTS  OF  PLANE  GEOMETRY. 


PAEALLELOGRAMS. 


THEOREM  XL. 

128.  In  any  parallelogram,  the  opposite  sides  and  angles  are 
equal. 

Let  ABCD  be  any  CJ. 

C  D 


r^"-  71     \ 

h 

c 

/      ^^^-^ 

/                                   ^-^                    / 

/                               "^^        J 

la                                       >«^-.^/ 

A 

B 

To  prove  that  AB  =  CD,  AC  =  BD,  Z.  a  =  Z.  ^>,  and 
Z_  ACD=  Z^ABD. 

Draw  the  diagonal  BC. 

Z_  7/1  =  Z_  n,  (62) 

^  c   =1.  d,  (62) 

and  BC=BC; 

A  ABC=  ^  CDB,  (84) 

and     AB=CD,AC=  BD,  and  Z^  a  =  Z.  b. 
From  the  first  two  equations,  ' 

Z^  c  -i-  Z^  n  =  Z^  d  -\-  ZL  m, 
or  Z,  ACD^  Z.  ABD.  Q. E. D. 

129.  CoK.  1. — A   diagonal  of  a  parallelogram  divides  it 
into  two  equal  triangles. 

130.  Cor.  2. — Parallels  intercepted   between  parallels   are 
equal. 


ELEMENTS  OF  PLANE  GEOMETRY.  61 


THEOREM  XLI. 

131.  Two  parallelograms  are  equal  if  two  sides  and  the 
included  angle  of  the  one  are  respectively  equal  to  two  sides 
and  the  included  angle  of  the  other. 

Let  ABCD  and  EFGH  be  two  zZ7s,  having  AD  =  EH, 
AB  =  EF,  and  Z.  a  =  Z.  6. 


C    H 


B      E 


To  prove  that  OJ  ABCD  =  CJ  EFOH. 

Draw  the  diagonals  BD  and  FH. 

AD  =  EH,  (Hyp.) 

AB  =  EF,  (Hyp.) 

and                                  Z.  a=:  Z.  b;  (Hyp.) 

A  ABD  =  A  EFH  (83) 

But  these  As  are  the  halves  of  the  zZ7s. 

CD  ABCD  =  cj  EFGH  Q.  E.  D. 


62  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XLII. 

132.  If  the  opposite  sides  of  a  quadrilateral  are  equal,  the 
figure  is  a  parallelogram. 

In  the  quadrilateral  ABCD,  let  AB  =  CD,  and  A  C=  BD. 
C  D 


a-  ^i 


A  B 

To  prove  that  ABCD  is  a  CJ. 
Draw  the  diagonal  B  C. 

AB  =  CD,  (Hyp.) 

AC=BD,  (Hyp.) 

and  BC=BC; 

AABC=^GDB,  (86) 

and  Z.  a  ^  Z.  b; 

^J5  and  Ci)  are  11.  (64) 

Also,  Z.  o  =  Z-  d; 

AC  and  BD  are  W,  (64) 

and  ABCD'm&CJ.  (121)      Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY.  63 


THEOREM  XLIII. 

133.  If  two  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  figure  is  a  parallelogram. 

Let  AB  CD  be  a  quadrilateral,  having  AB  and  CD  equal 

and  II. 


r\  ■ 

/    \^j 

A            *                                 B 

To  prove  that  AB  CD  is  a  CD. 

Draw  the  diagonal  BC. 

AB  =  CD, 

(Hyp.) 

BC=BC, 

and                                 JL  a=^  /-h; 

(62) 

AABC=^CDB, 

(83) 

and                                    /^  e=  Z.  d; 

^Cand^Darell, 

(64) 

and                                ^^ODlsao. 

(121) 

Q.E.D. 

64 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    XLIV, 

134.   The  diagonals  of  a  parallelogram  bisect  each  other. 
Let  ABCD  be  a  OJ,  and  AD,  B  C,  its  diagonals. 
C  D 


A  B 

To  prove  that  A  0  =  DO,  and  BO  =  CO. 

/^  a  =^  /-  c, 

l^h=  Z.d, 

and  AB=CD; 

AABO  =  ^CDO; 
whence        AO  =  DO,  and  BO  =  CO. 


(62) 

(62) 

(128) 

(84) 

Q.  E.  D. 


135.  Cor. —  The  diagonals  of  a  rhombus  bisect  each  other 
at  right  angles. 


ELEMENTS  OF  PLANE  GEOMETRY. 


65 


THEOREM  XLV. 

136.   The  diagonals  of  a  rectangle  are  equal. 

Let  AB  CD  be  a  rectangle,  and  J.Z>,  B  Q  its  diagonals. 


and 


and 


(128) 

(122) 

(83) 

Q.  E.  D. 


13Y.  Cor. — The  diagonals  of  a  square  are  eqxial  and  bisect 
each  other  ai  right  angles. 


oQ  ELEMENTS  OF  PLANE  GEOMETRY. 

THEOREM  XLVI. 

138.  If  a  parallel  to  the  base  of  a  triangle  bisects  one  of  the 
sides,  it  bisects  the  other  also;  and  the  part  of  it  intercepted 
between  the  sides  equals  half  the  base. 

In  the  A  J.^0,  let  DE  be  II  to  the  base  ^5  and  bisect 
ACoXD. 

C 


A  F 

I.  To  prove  that  DE  bisects  BC. 

Ijet  DF  he  W  to  B  a 

Z.  a=  Z-  c,  (62) 

Z.b=  Z.d,  (62) 

and  AD  =  DC;  (Hyp.) 

AADF=  A  DCE,  (84) 

and  DF  =  CE. 

FBDE  is  a  CJ;  (121) 

DF=EB. 
But  DF=  CE; 

EB=CE. 

II.  To  prove  that  DE  =  i  AB. 

In  the  equal  As  J.i>Fand  DCE, 
AF=DE, 
and  in  the  nj  FBDE, 

DE=FB;  (128) 

AF=  FB,  or  F  is  the  middle  point  of  AB. 
DE:=hAB.  Q.E.D. 

139.  Cor. — The  straight  line  which  joins  the  middle  points 
of  two  sides  of  a  triangle  is  parallel  to  the  third  side  and  is 
equal  to  half  that  side. 


ELEMENTS  OF  PLANE  GEOMETRY.  67 


THEOREM    XLVII. 

140.   The  parallel  to  the  bases  of  a  trapezoid,  bisecting  one  of 
the  non-parallel  sides,  bisects  the  other  also. 

Let  ABCD  be  a  trapezoid,  AB  and  DC  its  bases,  E  the 
middle  point  oi  AD,  and  let  EF  be  II  to  AB  and  DC. 


"---.0 


A  B 

To  prove  that  EF  bisects  B  C. 

Draw  the  diagonal  BD. 

In  the  A  ABD,  BD  is  bisected  at  0.  (138) 

Then  in  the  A  DBC,BC  is  bisected  at  F;  (138) 

^F  bisects^  a  Q.E.D. 

141.  Cor.  T/ie  straight  line  joining  the  middle  points  of  the 
non-parallel  sides  of  a  trapezoid,  is  parallel  to  the  base  and 
equals  half  their  sum. 


68 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM   XLVIII. 

142.  The  straight  line  drawn  from  the  vertex  of  a  right  angle 
of  a  right-angled  triangle  to  the  middle  of  the  hypothenuse  is 
equal  to  half  the  hypothenuse. 

Let  the  RA  ABC  be  right-angled  at  B,  and  let  BD  be 
drawn  to  the  middle  of  A  C. 


A  E 

To  prove  that  BD  =  ^  AC. 

Let  ED  be  II  to  BC. 

Z_  6  is  a  L. 

ED  bisects  AB; 

or,  AE  =  BE, 

ED  =  ED, 

and  Z-  a  =  /-  b ; 

aaed  =  abed, 

and  BD  =  AD=^AC. 


(63) 

(138) 

(45) 

(83) 

Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


69 


THEOREM  XLIX. 

143."  The  perpendiculars  drawn  from  the  vertices  of  a  triangle 
to  the  opposite  sides  meet  in  a  commo7i  point. 


In  the  A  ABC,  let  AD,  BE,  and  CF  be  the 
vertices  to  the  opposite  sides. 


from  the 


To  prove  that  AD,  BE,  and  CF  meet  in  a  common  point. 

Through  the  vertices,  let  MR,  GM,  and  GH  be  drawn 
respectively  11  to  AB,  BC,  and  AC. 

ABHC and  ABCMave  CDs;  (121) 

AB=CE:==MC,  (128) 

and  C  is  the  middle  point  of  MH. 

Kow  CF  is  _L  to  AB  and  MH;  (63) 

CF  is  _L  to  3fH  at  its  middle  point. 

Likewise  we  can  prove  that  AD  and  BE  are  J_s  to  GM 
and  GH  at  their  middle  points ; 

.*.     the  three  -Ls  meet  in  a  common  point.  (105)     Q.E. D. 


70  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    L. 

144.   The  medial  lines  of  a  triangle  meet  in  a  common  point. 
In  the  A  ABC,  let  AD,  BE,  and  CF  be  the  medial  lines. 
C 


To  prove  that  AD,  BE,  and  CF  meet  in  a  common  point. 

Let  AD  and  CF  meet  at  P,  and  let  M  and  N  be  the  middle 
points  of  CP  and  AP. 

Draw  MN,  NF,  FD,  MD. 

MN is  II  to  ^ 0 and  equals  h  AC,  (139) 

FD  is  I!  to  ^  C  and  equals  I  A  C;  (139) 

MN  and  FD  are  li  (66)  and  equal, 

and  NFDM  is  ncj;  (133) 

PF=  PM  (134)   =MC=^  CF. 

Or  AD  intersects  CF  at  P,  a  point  whose  distance  from  F 
equals  J  OP. 

Likewise  we  can  prove  that  BE  intersects  CF  at  a  point 
whose  distance  from  F  equals  i  CF. 

AD,  BE,  and  CF  meet  in  a  common  point.       Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY.  71 


EXEECISES  11^  IlSrYENTIOK 


THEOREMS. 

1.  The  two  straight  lines  which  bisect  the  two  pairs  of 
vertical  angles  formed  by  two  lines  are  perpendicular  to  each 
other. 

2.  Two  equal  straight  lines  drawn  from  a  point  to  a  straight 
line  make  equal  angles  with  that  line. 

3.  If  the  three  sides  of  an  equilateral  triangle  are  produced, 
all  the  external  acute  angles  are  equal,  and  all  the  obtuse 
angles  are  equal. 

4.  If  the  equal  angles  of  an  isosceles  triangle  are  bisected, 
the  triangle  formed  by  the  bisectors  and  the  base  is  an  isos- 
celes triangle. 

5.  The  three  straight  Imes  joining  the  middle  points  of  the 
sides  of  a  triangle  divide  the  triangle  into  four  equal  tri- 
angles. 

6.  If  one  of  the  acute  angles  of  a  right-angled  triangle  is 
double  the  other,  the  hypothenuse  is  double  the  shortest  side. 

7.  If  through  any  point  in  the  base  of  an  isosceles  triangle 
parallels  to  the  equal  sides  are  drawn,  a  parallelogram  is 
formed  whose  perimeter  equals  the  sum  of  the  equal  sides  of 
the  triangle. 

8.  If  the  diagonals  of  a  parallelogram  bisect  each  other  at 
right  angles,  the  figure  is  either  a  square  or  a  rhombus. 


72       ELEMENTS  OF  PLANE  GEOMETRY. 

9.  The  sum  of  the  four  lines  drawn  to  the  vertices  of  any- 
quadrilateral  from  any  point  except  the  intersection  of  the 
diagonals,  is  greater  than  the  sum  of  the  diagonals. 

10.  The  straight  lines  which  join  the  middle  points  of  the 
adjacent  sides  of  any  quadrilateral,  form  a  parallelogram 
whose  perimeter  is  equal  to  the  sum  of  the  diagonals  of  the 
given  quadrilateral. 

11.  Lines  joining  the  middle  points  of  the  opposite  sides  of 
any  quadrilateral,  bisect  each  other. 

12.  If  the  four  angles  of  a  quadrilateral  are  bisected,  the 
bisectors  form  a  second  quadrilateral  whose  opposite  angles 
are  supplements  of  each  other. 

Note. — If  the  figure  is  a  rhombus  or  a  square,  there  is  no  second  one 
formed. 


ELEMENTS  OF  PLANE  QEOMETBY.  73 


PROBLEMS. 


1.  Show  by  a  diagram  that  between  five  points,  no  three  of 

5X4 
which  lie  in  the  same  straight  line,  — ^ —  straight  lines  can 

be  drawn  connecting  the  points. 

2.  Between  n  points,  no  three  of  which  lie  in  the  same 
straight  line, -^ straight  lines  can  be  drawn  con- 
necting the  points. 

3.  What  is  the  greatest  number  of  angles  that  can  be 
formed  with  four  straight  lines  ?  Ans.  24. 

4.  If  the  sum  of  the  interior  angles  of  a  ^polygon  equals 
the  sum  of  its  exterior  angles,  how  many  sides  has  the 
polygon? 

5.  If  the  sum  of  the  interior  angles  of  a  polygon  is 
double  the  sum  of  its  exterior  angles,  how  many  sides  has  the 
figure  ? 

6.  If  the  sum  of  the  exterior  angles  of  a  polygon  is 
double  the  sum  of  its  interior  angles,  how  many  sides  has  the 
figure  ? 


74  ELEMENTS  OF  PLANE  GEOMETRY. 


BOOK  II. 

KATio  a:n^d  propoetiok 


DEFINITIONS. 

145.  To  measure  a  quantity  is  to  find  how  many  times  it 
contains  some  other  quantity  of  the  same  kind  called  the  unit 
of  measure. 

146.  Two  quantities  are  commensurable  if  they  have  a 
common  unit  of  measure. 

Two  quantities  are  incommensurable  if  they  have  no  com- 
mon unit  of  measure. 

Any  two  similar  quantities  may  be  considered  as  having  a 
common  unit  of  measure  infinitely  small. 

147.  In  Geometry  we  compare  two  similar  quantities  by 
finding  how  many  times  one  contains  the  other ;  that  is,  we 
measure  one  by  the  other.  The  magnitude,  therefore,  of 
a  quantity  is  always  relative  to  the  magnitude  of  some  other 
similar  quantity. 

148.  Ratio  is  the  measure  of  relation  between  two  similar 
quantities,  and  is  expressed  by  the  quotient  resulting  from 
dividing  the  first  by  the  second. 

The  first  of  the  two  quantities  compared  is  called  the 
Antecedent,  and  the  second  the  Consequent  Taken  together 
they  are  called  the  Terms  of  the  Ratio,  or  a  Couplet. 

Ratio  is  indicated  by  a  colon  placed  between  the  quantities 
compared,  or  by  the  fractional  form  of  indicating  division ; 

thus,  the  ratio  of  a  to  6  is  written,  a  :  6,  or  -y- 


ELEMENTS  OF  PLANE  GEOMETRY,  75 

149.  A  rroportion  is  an  expression  of  equality  between 
two  equal  ratios. 

Thus,  -r  =  -T.     This  means  that  the  ratio  of  a  to  6  equals 

the  ratio  of  c  to  d.     Usually  the  proportion  is  indicated  by  a 
double  colon  placed  between  the  two  couplets,  thus: 

a  :  b  ::  c  :  d. 

This  is  read,  a  is  to  b  as  c  is  to  d;  or,  the  ratio  of  a  to  6 
is  equal  to  the  ratio  of  c  to  d. 

Of  the  four  terms  compared,  the  first  and  third  are  the 
Antecedents,  and  the  second  and  fourth  are  the  Consequents. 
The  Extremes  are  the  first  and  fourth  terms.  The  Means 
are  the  second  and  third  terms.  The  Fourth  Proportional 
is  the  fourth  term.    When  the  means  are  equal,  as  in 

a  \   b    '.'.    b   '.   c, 

b  is  said  to  be  the  Mean  Proportional  between  a  and  c ;  and  c 
is  said  to  be  the  Third  Proportional  to  a  and  b. 

150.  Four  quantities  are  Reciprocally  Proportional  when 
the  first  is  to  the  second  as  the  reciprocal  of  the  third  is  to 
the  reciprocal  of  the  fourth. 

Thus,  a  :  6  ::—:-,  • 

c       a 

Two  quantities  and  their  reciprocals  form  a  reciprocal 
proportional. 

Thus,  a  :  b  ::  t-  :  — 

6      a 

151.  A  proportion  is  takenby  JLftema^ion,  when  antecedent 
is  compared  with  antecedent,  and  consequent  with  consequent. 

Thus,  if  a  :  6  : :  c  :  6?,  we  have  by  alternation  either 
a  I  G  ::   b  i  d;  ov,  d  :  b   ::   c  :  a. 


76       ELEMENTS  OF  PLANE  GEOMETRY. 

152.  A  proportion  is  taken  by  Inversion^  when  the  ante- 
cedents are  made  consequents  and  the  consequents  antece- 
dents. 

Thus,  if  a  :  6  : :  c  :  cf ,  we  have  by  inversion  h  :  a  : :  d  :  c. 

153.  A  proportion  is  taken  by  Composition,  when  the  sum 
of  antecedent  and  consequent  is  compared  with  either  ante- 
cedent or  consequent. 

Thus,  if  a  :  h  ::  c  :  c^,  we  have  by  composition 

a  -\-  h  :  a  ::   c  -{-  d  :  c; 

or  a  -\-  h   '.   b    '.'.    c  -\-  d  '.   d. 

154.  A  proportion  is  taken  by  Dimsion,  when  the  difference 
of  antecedent  and  consequent  is  compared  with  either  ante- 
cedent or  consequent. 

Thus,  if  a  :   h   ::  c  :   c^,  we  have  by  division 

a  —  h  :   a   ::    c  —  d  :  c; 

or  a  —  h  :   h   '.'.   c  —  did. 

155.  A  Continued  rroportion  is  a  series  of  equal  ratios. 

Thus,  \i  a  '.  h  ::  h  :.  c  ::  c  :  d  ::  d  :  e,  we  have  a 
continued  proportion. 


ELEMENTS  OF  PLANE  GEOMETRY.  77 


THEOREM    I. 

156.  In  any  proportion  the  product  of  the  extremes  is  equal 
to  the  product  of  the  means. 

Let  a  :  b   ::  G  :  d. 

To  prove  that    a  d  =  b  c. 

ft         /» 

Take  the  form  r  =  -r- 

Multiply  both  members  by  b  d; 
then  a  d  ==b  c.  Q.  E.  D. 


THEOREM   II. 

157.  If  the  product  of  two  quantities  equals  the  product  of 
two  others,  the  quantities  in  either  product  can  be  made  the 
means,  and  those  of  the  other  product  the  extremes  of  a  propor- 
tion. 

Let  a  d  =  b  c. 

To  prove  that    a  :  b   ::   c  :  d.  • 

Divide  both  members  by  b  d. 


Then  4=4. 

b  d 

or  a  :  b  ::  c  :  d.  Q.  E.  D. 

7* 


78  ELEMENTS  OF  PLANE  GEOMETRY. 

THEOREM    III. 

158.  A  mean  proportional  between  two  quantities  equals  the 
square  root  of  their  product. 

Let  a  :  h   ::   h  :  c. 

To  prove  that    h  =  \/  a  c,  ^ 

b'  =  ac.  (156) 

Extract  the  square  root  of  both  members. 

Then  b  =  ^/Ta  Q.  E.  D. 


THEOREM    IV. 

159.   The  corresponding  members  of  two  equations  form  the 
couplets  of  a  proportion. 


Let 

a  =  c, 

and 

b=:d. 

To  prove  that    a  : 

b  ::  e:  d. 

Divide. 

Then 

a           c 

or  a  :  b  ::  c  :  d.  Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY.  79 


THEOREM  V. 

160.  If  four  quantities  are  in  proportion,  they  are  in  pro- 
portion hy  alternation. 

Let  a  \  b   ::  c  :  d. 

To  prove  that    a  :  c  ::   b  :  d. 

Take  the  form  -—  =  —-. 

6        d 

Multiply  both  members  by  — 

Then  ^=A, 

c        a 

or  a  :  c  ::   b  :  d.  Q.E.D. 


THEOREM  VI. 

161.  If  four  quantities  are  in  proportion^  they  are  in  pro- 
portion by  inversion. 

Let  a  :  b   w   c  '.  d. 

To  prove  that    b  :  a  ::  d  :  e. 

Take  the  form  —  =  —-. 

0        a 

Divide  1  by  each  member. 

Then  ^  =  1, 

a         e 

or  b  :  a  ::   d  :  c,  Q.E.D. 


80  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  VII. 

162.  If  Jour  quantities  are  in  proportion^  they  are  in  pro- 
portion by  composition. 

Let  a  :  b  ::  c  \  d. 

To  prove  that  a  -\-  b  :  b  ::  c  -\-  d  :  d. 

Take  the  form  t~  =  ~r* 

b        a 

To  each  member  add  1. 

Then  |.4-i:^-|.4_l; 


whence 


a  -\-  b  c  -{-  d 


b  d 

or  a^b  -.b  ::  G-\-  d  \  d.  Q.  E.  D. 

THEOREM  VIII. 

163.  If  Jour  quantities  are  in  proportion,  they  are  in  pro- 
portion by  division. 

Let  a  :  b  w  G  :  d. 

To  prove  that    a  —  b  :  b  ::  c  —  d  :  d. 


Take  the  form 

a 

c 
~d' 

From  each  member  subtract  1. 

Then 

a 
T 

—  1  = 

c 

-1; 

bence 

a  — 
b 

_6  _c 

—  d 

d     ' 

a  —  b  : 

b  ::  c- 

-d  : 

d. 

or  a  —  b  :  b  ::  c  —  d  :  d.  Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY.  81 


THEOREM    IX. 

164.  If  two  proportions  have  a  couplet  in  each  the  same,  the 
other  couplets  form  a  proportion. 


Let                            a  :  b  ::  G  :  d, 

and                               a  :  b  ::  e  :  f. 

To  prove  that    c  :  d  ::  e  :  f. 

Take  the  forms  --  =  -5-  and  -y-  = 
0          a            0 

e 
7* 

Then                              4  =  4-» 

(Ax.  1.) 

or                                  c  :  d  ::  e  :  f. 

Q.E.D. 

THEOREM  X. 

165.  Equimultiples  of  two  quantities  are  proportional  to  the 
quantities  themselves. 

Let  a  and  b  be  any  two  quantities. 

To  prove  that  ma  :  mb  ::  a  :  b. 

a  a 

Multiply  both  terms  of  the  first  member  by  m. 

Then  '^  =  -%, 

mb         0 

or  ma  :  mb  ::  a  :  h  Q. E. D. 


82  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XI. 

166.  In  any  proportiony  any  equimultiples  of  the  first  couplet 
are  proportional  to  any  equimultiples  of  the  second  couplet. 

Let  a  :  b   ::   c  :  d. 

To  prove  that    ma  :  mb   ::   nc  :  nd. 

Take  the  form  -y-  =  ^r* 

0         a 

Multiply  both  terms  of  the  first  member  by  m,  and  both 
terms  of  the  second  member  by  n. 


Then 


ma 7ic 

mb        nd 


or 


ma  :  mb   ::   nc  :  7id.  Q.  E.  D. 


THEOREM  XII. 


167.  If  two  quantities  are  increased  or  diminished  by  like 
parts  of  each,  the  results  are  proportional  to  the  quantities 
themselves. 

Let  a  and  b  be  any  two  quantities. 

To  prove  that  a  ±.  —  a  :  b  ±  —  b   ::   a  :  b. 

ma  :  mb   ::   a  :  b.  (165) 

For  m,  substitute  1  ±:  —  • 

Then       (l  ±  ^^   a  :    (l  it  ^)   b    ::   a  :  b, 


q/  \  q 

P-a:b±P- 


or  a±  ^  a  :  b  ±^   b  ::  a  :  b.       Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY.  83 

THEOREM   XIII. 

16.8.  In  any  continued  proportion,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 

Let  a  :  h   :'.   c  :  d  w   e  :  f  '.:  g  :  h. 

To  prove  that  a  -\-  c  -\-  e  -\-  g  :   b  -{-  d  -{-  f  -\-  h  ::  a  :  b. 

Denote  the  common  ratio  by  r. 

mi  a  c  e  a 

Then  r=-^=^^  =  j  =  JL. 

Whence        a  =  br,   c  =  dr,   e  =  fr,   g  =  hr. 
Add  these  equals. 

Then    a  +  c  +  e  +  ^  =  (64-^+/+/i)r. 
Divide  both  members  hy  (b-\-d-\-f-^h). 
rru^^  a  +  c-j-  e  +  g  a 

b  -{-  d  +/+/1  b 

or         a  4-  c  +  e  +  ^  :  6  +  ^  +/  -f  /i  ::  a  :  6.    Q.  E.  D, 

THEOREM  XIV. 

169.  In  two  or  more  proportions,  the  products  of  the  corre- 
sponding terms  are  proportional. 

C a  :  b   \:  c  :  d, 

Let  <^  e  :  f  ::  g  :  h,  ^ 

\m  '.  n  w   0  :  p. 

To  prove  that  aem  :  bfn   : :   ego  :  dhp. 

Take  the  forms  -^  =  — r?  --r  =  4-^   —  =  — 
0  d      J  h      n  p 

By  multiplication  we  have 

aem  ego 

bfn         dhp 
or  aem  :  bfn   : :   ego  :  dhp.  Q.  E.  D. 


84  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    XV. 

170.  Like  powers,  or  like  roots,  of  the  terms  of  a  proportion 
form  a  proportion. 

Let  a  :  h   ::   c  :  d. 

To  prove  that    aJ"  :   6"  ::   c"  :   d", 
and  that  a"  :   6"  : :   c"  :   d~. 

Take  the  form  -7-  =  — 7- 

0  a 

Raise  each  member  to  the  ?i'*  power. 

Then  4l  =  4"' 

V        d" 

or  a"  :  6"  ::   c"  :  c?". 

Also  extract  the  ?i'*  root  of  each  member.  ' 

Then  —  ==  — ' 

b^        d-^ 

or  a"  :   6"  : :   c"  :   d».  Q.  E.  D. 

171.  Scholium.     The  product  of  two  quantities  implies 
that  at  least  one  is  numerical. 

In  (169)  and  (170),  all  the  quantities  must  be  numerical. 

In  (160)  and  (168),  all  the  quantities  must  be  of  the  same 
kind. 


ELEMENTS  OF  PLANE  GEOMETRY.  85 


BOOK  III. 

THE    CIRCLE. 


DEFINITIONS. 

172.  A  Circle  is  a  plane   bounded  by  a  curve,  all   the 
points  of  which  are  equally  distant  from  q^ 
a  point  within,  called  the  centre. 

A\ 
The  Circumference  of  a  circle  is  the 
curve  which  bounds  it. 

An  Arc  is  a  part  of  the  circumference ;  as,  ^  C.  A  Semi- 
circumference  is  an  arc  equal  to  half  of  the  circumference. 

A  Radius  is  a  straight  line  extending  from  the  centre  to 
any  point  in  the  circumference;  as,  OC. 

173.  A  Diameter  of  a  circle  is  a  straight  line  passing 
through  the  centre  and  terminating  each  way  in  the  circum- 
ference; as,  AB. 

174.  A  Chord  is  a  straight  line  joining 
any  two  points  in  the  circumference;   as,      P/ 

^^'  JD^ '0 

The  arc  EPD  is  said  to  be  subtended  by 
its  chord  ED.     Every  chord  subtends  two 
arcs,  whose  sum  equals  the  whole  circumference.     Whenever 
an  arc  and  its  chord  are  spoken  of,  the  less  arc  is  meant. 

8 


8Q 


ELEMENTS  OF  PLANE  GEOMETRY. 


'  175.  A  Segment  of  a  circle  is  the  portion  enclosed  by  an 
arc  and  its  chord;  as,  FGH.  A  Semicircle  is  a  segment 
equal  to  one-half  of  the  circle. 

176.  A  Sector  of  a  circle  is  the  por- 
tion enclosed  by  an  arc  and  the  radii 
drawn  to  its  extremities ;  as,  OFM. 

111.  A  Tangent  is  a  straight  line 
which  touches  the  circumference  but 
does  not  intersect  it;  as,  ABC. 

The  common  point  B  is  called  the 
point  of  contact,  or  the  point  of  tangency. 

178.  A  Secant  is  a  straight  line  which 
cuts  the  circumference  in  two  points ;  as,  DE. 

179.  An  Inscribed  Angle  is  one  whose 
vertex  is  in  the  circumference  and  whose 
sides  are  chords;  as,  ABE. 

180.  An    Inscribed,    Polygon  .  is     one 

whose  sides  are   chords  of  a   circle;  as,  B^  D 

ABCDEF.     The   circle   is   then   said   to   be   circurtiscrihed 
about  the  polygon. 

181.  A  Polygon  is  circumscribed 
about  a  circle  when  all  its  sides  are 
tangents  to  the  circumference;  as, 
MNOPQ.  The  circle  is  then  said 
to  be  inscribed. 

182.  By  the   definition   of  a   circle, 
all   its   radii    are    equal;    also,  all    its  Q 
diameters  are  equal.     It  also  follows  from  the  definition  that 
circles  are  equal  when  their  radii  are  equal. 


ELEMENTS  OF  PLANE  GEOMETRY.  87 


CHOEDS,  ARCS,  AISTGLES  AT  THE 
CENTRE,  SECANTS,  AND  RADII. 


THEOREM  I. 

183.  Any  diameter  bisects  the  circle  and  its  drcumference. 
Let  ABCD  be  a  O,  and  AB  any  diameter. 

C 


D 

To  prove  that  ABC=  ABD. 

On  AB  as  an  axis,  revolve  the  portion  ABC  till  it  falls  in 
the  plane  of -4-Si). 

Then  the  curve  ACB  coincides  with  the  curve  ADB,  for 
all  the  points  in  each  are  equally  distant  from  the  centre  0. 

ABC  =  ABD,  (14) 

and  curve  ACB  =  curve  ADB.  Q. E.  D. 


88  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    II. 

184.  A  diameter  of  a  circle  is  greater  than  any  other  chord. 


In   the   O  ACB,  let  AB   he  any  diameter  and  BC  any 
other  chord. 


ToprovethatAB>  BC. 

From  the  centre  0,  draw  OC,  ' 

0A=  Oa  (182) 

0C-\-  0B>  BC.  (Ax.  11) 

Substitute  OA  for  its  equal  OC. 
Then  0A-\-  0B>  BC, 

or  AB>Ba  (Q.E.D.) 


ELEMENTS  OF  PLANE  GEOMETRY. 


89 


THEOREM  III. 

185.  A  straight  line  cannot  cut  the  circumference  of  a  circle 
in  more  than  two  points. 

In  the  O  ADBC,  let  AB  cut  the   circumference  at  A 
and  B. 


To  prove  that  A  B  cannot  cut  the  circumference  in  more  than 
two  points. 


Draw  the  radii  OA,  OB,  OD, 

0A=  0B=  OD. 


(182) 


li  AB  could  cut  the  circumference  at  A,  B,  and  2),  there 
would  be  three  equal  straight  lines  drawn  from  the  same 
point  to  the  same  straight  line,  which  cannot  be.  (61) 

.  • .  a  straight  line  cannot  cut  a  circumference  in  more  than 
two  points.  Q.  E.  D. 

8* 


90 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM   IV. 

186.  In  equal  circles,  or  in  the  same  circle,  equal  angles  at 
the  centre  iyitercept  equal  arcs. 

Let  0  and  P  be  the  centres  of  the  equal  Os  ABC  and 
DEF,  and  let  Z.  a  =  ^  5. 


To  prove  that  arc  AB  =  arc  DE. 

Place  the  OABC  on  the  ODEF  so  that  Z_  a  coincides 
with  Z_  h. 

0A  =  OB  =  PD  =  PE;  (182) 
A  falls  on  D,  and  B  falls  on  E; 

ABO  =  DEP,  (14) 

and                               arc  AB  =  arc  DE.  Q.  E.  D. 


187.  Cor. — Li  equal  circles,  or  in  the  sayne  circle,  equal  arcs 
subtend  equal  angles  at  the  centre. 


ELEMENTS  OF  PLANE  GEOMETRY. 


91 


THEOREM  V. 

188.  In  equal  circles,  or  in  the  same  circle,  equal  arcs  are 
subtended  by  equal  chords. 

In  the  equal  Os  ABE  and  CDF,  let  arc  AB  =  arc  CD, 


E  F 

To  prove  that  chord  AB  =  chord  CD. 
Draw  the  radii  OA,  OB,  PC,  PD. 

OA  =  PC,  (182) 

OB  =  PD,  (182) 

and  Z.a=/-b;  (187) 

AABO  =  ACDP,  (83) 

and  chord  AB  =  chord  CD.  Q.  E.  D. 


189.  Cor. — In   equal  circles,  or  in  the  same  circle,  equal 
chords  subtend  equal  arcs. 


92 


ELEMENTS  OF  PLANE  GEOMETRY, 


THEOREM   VI. 

190.   TJie  radius  petpeyidicular  to  a  chord  bisects  the  chord 
and  the  subtended  arc. 

Let  OCbe  a  radius  _L  to  the  chord  AB  at  D. 


To  prove  that  AD  =  BD,  and  arc  AC  =  arc  BC. 
Draw  the  radii  OA  and  OB. 


and 

and 
Also 


0A=  OB,  (182) 

OD  =  OD; 
RA  AOD  =  KA  BOD,  (91) 

AD  =  BD. 
Z.  a=  ^  b; 
2irG  AC  =  fiYG  BC.        (186)     Q.  E.  D. 


191.  Cor. — The  perpendicular  erected  at  the  middle  point 
of  a  chord  passes  through  the  centre  of  the  circle  and  bisects 
the  subtended  arc. 


ELEMENTS  OF  PLANE  GEOMETRY.  93 


THEOREM  VII. 

192.  In  the  same  circle,  or  in  equal  circles,  equal  chords  are 
equally  distant  from  the  centre;  and  if  two  chords  are  unequal, 
the  less  is  at  the  greater  distance  from  the  centre. 

Let  chord  AB  =  chord  CD,  and  chord  CE  <  chord  CD; 
and  let  OF,  0  G,  and  OH  be  _Ls  to  these  chords  from  the 
centre  0. 


To  prove  that  OF=OG,  and  OH  >  OG. 

OF  and  OG  bisect  the  equal  chords ^5  and  CD;      (190) 
AF=  CG; 
'RAAOF='RA  COG,  (91) 

and  0F=  OG. 

Again,  CD  >  CE; 

OH  cuts  CD  in  some  point,  as  P. 
Now  0H>  OP. 

But  0P>  OG;  (62) 

still  more  is  OH  >  0  G.  Q.  E.  D. 


94  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM   VIH. 

193.   Through  three  points  not  in  the  same  straight  line,  a 
circumference  of  a  circle  can  be  passed. 

Let  A,  B,  and   C  be  any  three  points  not  in  the  same 
straight  line. 


To  prove  that  through  A,  B,  and  C,  a  circumference  of  a  O 
can  he  passed. 

Draw  AB  and  BC,  and  at  their  middle  points  let  the  -Ls 
EM  and  DP  be  erected. 

AB  and  B  C  are  not  in  the  same  straight  line ; 

the  _J_s  EM  and  DP  meet  in  some  point,  as  O. 

Now,  0  is    equally  distant  from  A   and    B;    also   from 
B  and  C;  (53) 

.  • .  O  is  equally  distant  from  A,  B,  and  C,  and  a  circum- 
ference with  OA  as  a  radius  passes  through  these  points. 

Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


95 


THEOREM    rx,,  .e-       **X] 

194.  A  straight  line  perpendicular  to  a  radimlit^  extrehiity 
is  a  tangent  to  the  circle. 

Let  AB  be  J_  to  the  radius  OP  at  P. 


To  prove  that  AB  is  a  tangent  to  the  O  a^  the  point  P. 

From  the  centre  draw  any  oblique  line,  as  OG 

0C>  OP;  (52) 

.  • .     the  point  C  is  without  the  circumference,  and  all  points 
in  AB,  except  P,  are  without  the  circumference; 

AB  is  a  tangent  to  the  O  at  P.     (177)     Q.  E.  D. 

195.  Cor. — A  straight  line  tangent  to  a  circle  is  perpen- 
dicular to  the  radius  drawn  to  the  point  of  contact. 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    X. 

196.  Two  parallel  secants  intercept  equal  arcs. 

Let  the  lis  AB  and  CD  intercept  the  arcs  ^C  and  BD. 


To  prove  that  arc  AC  =  arc  BD. 

Suppose  the  radius  OE  to  be  drawn  _L  to  AB  and  CD. 

Then  arc  AE  =  arc  BE, 

and  &TC  CE=  arc  DE.  (190) 

Subtract; 
then  arc  AE  —  arc  CE  =  arc  BE  —  arc  DE. 

or  arc  AC  =  arc  BD.  Q.  E.  D. 

197.  Scholium. — This  proposition  is  true  for  any  position 
of  tjie  parallels;  hence  it  is  true  if  one  or  both  become 
tangents ;  and  the  straight  line  which  joins  the  points  of 
contact  of  two  parallel  tangents  is  a  diameter. 


ELEMENTS  OF  PLANE  GEOMETRY.  97 


EELATIYE  POSITION   OF  CIECLES. 


THEOREM  XI. 

198.  If  two  circles  cut  each  other,  the  straight  line  joining 
their  centres  bisects  their  common  chord  at  right  angles. 

Let  AB  be  a  common   chord  of  two  O  s  which  cut  each 
other,  and  00  join  the  centres  0  and  O. 


To  prove  that  0  C  is  -L  to  AB  at  its  middle  point. 
Draw  the  radii  OA,  OB,  CA,  and  CB. 

OA  =  OB,  (182) 

and  CA  =  CB;  (182) 

0  is  equally  distant  from  A  and  B, 
and  Ois  equally  distant  from  A  and  B; 

OCis  J- to  AB  eit  its  middle  point.     (55)     Q.  E.  D. 

199.  Cor. — If  two  circles  touch  each  other,  either  externally  or 
internally,  the  point  of  contact  is  in  the  line  joining  their  centres. 


98       ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XII. 

200.  If  two  circles  cut  each  othevy  the  distance  between  their 
centres  is  less  than  the  sum  and  greater  than  the  difference  of 
their  radii. 

Let  0  and  C  be  the  centres  of  two  Os  whose  circumferences 
cut  each  other  at  A  and  B,  and  draw  the  radii  OA  and  CA. 


To  prove  that  0C<  OA  +  CA,  and  0C>  OA  —  CA. 
The  point  A  does  not  lie  in  OC;  (199) 

OC^  is  a  A. 
Now,  OC<OA-\-  CA,  (Ax.  11) 

and  0C>  OA—  CA.  (73)     Q.  E.  D. 

201.  Cor.  1. — If  two  circles  touch  each  other  externally,  the 
distance  between  their  centres  equals  the  sum  of  their  radii. 

202.  "Cor.  2. — If  two  circles  are  wholly  exterior  to  each 
other,  the  distance  between  their  centres  is  greater  than  the  sum 
of  their  radii. 


ELEMENTS  OF  PLANE  GEOMETRY. 


99 


THE  MEASUEEMEE-T  OF  AITGLES. 


THEOREM  XIII. 

203.  In  equal  circles,  or  in  the  same  circle,  angles  at  the 
centre  are  to  each  other  as  the  arcs  which  they  intercept. 

In  the  equal  Os  ABM  and  CDN,  let  0  and  P  be  the 
centres,  and  let  the  Z.s  AOB  and  CPD  intercept  the  arcs 
AB  and  CD. 

M  N 


To  prove  that  A.  AOB  :   Z.  CPD  ::  arc  AB  :  arc  CD. 

Let  EF  be  a  common  unit  of  measure  of  AB  and  CD, 
and  suppose  it  to  be  contained  in  AB  8  times,  and  in  CD 
5  times. 

Then  arc  AB  :  arc  CD  ::  S  :  5. 

Draw  radii  at  the  several  points  of  division  of  the  arcs. 

Then  the  partial  ZL  s  are  equal.  (187) 

AOB  contains  8,  and  CPD  contains  5  equal  Z-s; 

/LAOB  :   Z.  CPD  ::  8  :  5. 

But  arc^-B  :  arc  CD  ::  S  :  5; 

/.  AOB  :  Z.  CPD  ::  arc  AB  :  arc  CD. 


100  ELEMENTS  OF  PLANE  GEOMETRY. 

The  same  proportion  is  found  if  other  numbers  than  8  and 
5  are  taken. 

Now,  this  is  true,  whatever  may  be  the  length  of  the  com- 
mon unit  of  measure  of  the  arcs;  hence  it  is  true  when  it  is 
iniinitely  small,  as  is  the  case  when  the  arcs  are  incom- 
mensurable.    Hence,  in  any  case,  the  proposition  is  true. 

Q.E.D. 


204.  Cor. — In  equal  circles^  or  in  the  same  circle^  arcs  which 
are  intercepted  by  angles  at  the  centre  are  to  each  other  as  the 
angles. 

205.  Scholium. — The  truth  of  this  proposition  gave  rise  to 
the  method  of  measuring  angles  by  arcs.  It  will  be  observed, 
that  if  arcs  are  struck  with  the  same  radius  from  the  vertices 
of  angles  as  centres,  the  angles  are  to  each  other  as  the  arcs 
intercepted  by  their  sides.  Hence  the  angle  is  said  to  be 
measured  by  the  arc. 

The  unit  of  measure  generally  adopted  is  an  arc  equal  to 
3  J  (J  of  the  circumference  of  a  circle,  called  a  degree,  and 
denoted  thus  (°). 

The  degree  is  divided  into  60  equal  parts,  called  minutes, 
denoted  thus,  ('). 

The  minute  is  divided  into  60  equal  parts,  called  seconds, 
denoted  thus,  ("). 

A  right  angle,  therefore,  is  measured  by  90°;  or,  as  we  say, 
it  is  an  angle  of  90°. 

An  angle  of  45°  is  ^  of  a  right  angle ;  an  angle  of  30°  is 
i  of  a  right  angle.  Thus  we  have  a  definite  idea  of  the 
magnitude  of  an  angle  if  we  know  the  number  of  degrees  by 
which  it  is  measured. 


ELEMENTS  OF  PLANE  GEOMETRY. 


101 


THEOREM    XIV. 

206.  An  inscribed  angle  is  measured  by  half  the  intercepted 
arc. 

In  theO^-BC,   let  AB  and  CB  be  the  sides  of  the  in- 
scribed Z.  a. 

Case  I. — Let  the  centre  0  be  in  one  of  the  sides. 

B 


To  prove  that  Z_  a  is  measured  by  i  arc  A  C. 

Draw  the  radius  OA. 

OA  =  OB; 

(182) 

. 

^  ABO  is  isosceles, 

id 

z:a  =  Z.b. 

(94) 

Z-c  =  a  /-  -^  L.b. 

(82) 

Substitute 

Z.  a  for  its  equal  Z.  b. 

Then 

Z.  c  =  2  21  a.        .     - 

Now, 
id 

Z.  c  is  measured  by  arc  A  C; 

2  Z_  a  is  measured  by  arc  A  C, 

Z_  a  is  measured  by  2  A  C. 
9* 

(205) 

102 


ELEMENTS  OF  PLANE  GEOMETRY. 


Case  11.— Let  the  centre  0  fall  within  the  inscribed  Z_. 

B 


To  prove  that  Z_  ABC  is  measured  by  ^  arc  A C. 
Draw  the  diameter  BD. 

Z.  a  is  measured  by  ^  arc  AD, 
and  Z.  6  is  measured  by  ^  arc  D  C;  (Case  I.) 

Z^  a  -\-  /-  b\^  measured  by  ^  (arc  AD  +  arc  DC), 
or      Z.  ABC  is  measured  by  h  AC. 


Case  III. — Let  the  centre  0  fall  without  the  inscribed  Z_ . 

B 


To  prove  that  Z_  a  is  measured  by  i  arc  AC. 

Draw  the  diameter  BD. 

Z.  DBC  is  measured  by  ^  arc  DC, 
and  Z.  6  is  measured  by  i  arc  DA ;  (Case  I.) 

.  • .    Z.  D^C  —  Z_  6  is  measured  by  J  (arc  DC —  arc  DA), 
or  Z.  a  is  measured  by  ^  arc  J.  C  Q.  E.  D, 


ELEMENTS  OF  PLANE  GEOMETRY.  103 

.^ ft . 

207.  Cor.  1. — All  angles  inscribed  in  the  same  segment  are 
equal. 

208.  Cor.  2. — Any  angle  inscribed  in  a  semicircle  is  a 
right  angle. 

209.  Cor.  3. — Any  angle  inscribed  in  a  segment  less  than  a 
semicircle  is  obtuse. 

210.  Cor.  4. — Any  angle  inscribed  in  a  segment  greater 
than  a  semicircle  is  acute. 


104  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XV. 

211.  Any  angle  formed  by  a  tangent  and  a  chord  is  measured 
by  half  the  intercepted  arc. 

Let  the  Z.  ABD  be  formed  by  the  tangent  CD  and  the 
chord  AB. 


To  prove  that  Z.  ABD  is  measured  by  J  arc  AEB. 

Draw  the  diameter  BE. 
Z.  6  =  a  L,  (195)  and  is  measured  by  ^  arc  EB,  (208) 
and  Z.  a  is  measured  by  h  arc  AE;  (206) 

.  • .    Z-  a  +  Z_  6  is  measured  by  h  (arc  AE  -{-  arc  EB), 
or     Z.  ABD  is  measured  by  J  arc  AEB.  Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY.  105 


THEOREM    XVI. 

212.  Any  angle  formed  by  two  chords  intersecting  is  measured 
by  half  the  sum  of  the  arcs  intercepted  between  its  sides  and  the 
sides  of  its  vertical  angle. 

Let  the  Z.  a  be  formed  by  the  chords  AB  and  CD. 


To  prove  that  Z.  a  is  measured  by  ^  (arc  AD  -\-  arc  GB). 

DrawylC. 

Z.  c  is  measured  by  h  arc  AD, . 
and  Z.  6  is  measured  by  l  arc  CB.  (206) 

But  ^a=  /_  c^  l.h;  (82) 

Z.  a  is  measured  by  h  (arc  AD  +  arc  GB). 

Q.KD. 


106 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    XVH. 

213.     Any  angle  formed  by  two  secants  is  measured  by  half 
the  difference  of  the  intercepted  arcs. 

Let  the  Z.  a  be  formed  by  the  secants  AB  and  BC. 

B 


To  prove  that  /L  a  is  measured  by  I  {arc  AC —  arc  EF). 
Let  Z)£  be  II  to  O^. 

arc  DC  =  arc  EF.  (196) 

Z.  6  is  measured  by  i  arc  AD. 

But  arc  AD  =  (arc  AC  —  arGDC)  =  (arc  ^  O—-  arc  EF,) 

and  Z^b=  Z.  a;  (62) 

Z.  a  is  measured  by  i  (arc  AC  —  arc  EF).     Q. E.  D. 

214.  Scholium. — This  proposition  is  true  for  any  position 
of  the  secants;  hence  it  is  true  if  one  or  both  become 
tangents. 


ELEMENTS  OF  PLANE  GEOMETRY.  107 


PEOBLEMS  IN  COIsrSTEUCTIOK 


PROBLEM  I. 

215.  To  bisect  a  given  straight  line. 

Let  AB  be  the  given  straight  line. 


B 


/E\ 


With  A  and  B  as  centres  and  a  radius  greater  than  half 
of  AB,  describe  arcs  intersecting  at  C  and  E. 

Draw  CE. 

Then         CE  bisects  AB  at  the  point  P.     (55)  Q.E.F. 


108 


ELEMENTS  OF  PLANE  GEOMETRY, 


PROBLEM  n. 

216.  At  any  point  in  a  straight  line  to  erect  a  perpendicular. 
Let  P  be  any  point  in  the  straight  line  AB. 


D 


E 


Cut  off  PD  =  PE. 


With  D  and  E  as  centres  and  a  radius  greater  than  PD 
or  PE^  describe  arcs  intersecting  at  C 

Draw  CP. 


Then 


CPh  A- to  AB.  (55)    Q.E.F. 


ELEMENTS  OF  PLANE  GEOMETRY.  109 


PROBLEM   III. 

217.  From  any  point  without  a  straight  line  to  draw  a  per- 
pendicular to  that  line. 

Let  P  be  any  point  without  the  straight  line  AB. 


^    c--^,,.,.|.,.,.^--^ 


.^^^ 


With  P  as   a   centre    and   a  radius   sufficiently  great, 
describe  an  arc  cutting  AB  2X  C  and  D. 

With  C  and  B  as  centres  and  a  radius  greater  than  half 
of  CD^  describe  arcs  intersecting  at  E. 

Draw  FE. 

Then  isPjEJLto^^.  (65)    Q.E.F. 


10 


110 


ELEMENTS  OF  PLANE  GEOMETRY. 


PROBLEM    IV. 


218.   To  bisect  a  given  arc. 
Let  AOB  be  a  given  arc. 


/\ 


0  X./ 


♦  Draw  the  chord  AB. 

Bisect  AB  by  a  _1_  as  in  (215). 
This  _L  bisects  the  arc. 


(191)    Q.E.F. 


ELEMENTS  OF  PLANE  GEOMETRY.  Ill 


PROBLEM  V. 

219.  To  construct  an  angle  equal  to  a  given  angle^  at  any 
point  in  a  line. 

Let  a  be  the  given  /_ ,  and  A  the  point  in  the  line  AB. 


V 


A^^ L B 


With  the  vertex  0  as  a  centre,  and  any  radius,  describe 
an  arc  cutting  the  sides  of  Z.  a  at  if  and  N. 

With  ^  as  a  centre   and  the   same   radius,  describe  the 
indefinite  arc  CD. 

Draw  the  chord  M  N. 

With  D  as  a  centre  and  iltf  iV  as  a  radius,  describe  an  arc 
cutting  the  indefinite  arc  at  C. 

Drawee. 

Then  arc  CD  =  arc  M  N;  (189) 

/.h=l.a.  (187)    Q.E.F. 


112 


ELEMENTS  OF  fLANE  GEOMETRY, 


PROBLEM  VI. 

220.  To  bisect  a  given  angle. 
Let  ABC  hQ  the  given  Z.. 


With  J5  as  a  centre    and   any   radius,  describe  an  arc 
cutting  the  sides  of  the  Z.  at  i)  and  E. 

Bisect  the  arc  as  in  (218). 

Then,  since  arc  DF  =  arc  FE,  Z.  a=  ^h.  (187) 

Q.E.F. 


PROBLEM  VH. 

221.   Through  a  given  point,  to  draw  a  straight  line  parallel 
to  a  given  straight  line. 

Let  P  be  the  given  point,  and  AB  the  given  straight  line. 


D 


PA 

/ 

/a 


/b 


From  any  point  in  AB,  as  C,  draw  the  line  CD  through  P. 

At  P  construct  the  Z-  fe  =  Z.  a  as  in  (219). 
Then  P^  is  II  to  ^5.  (64)    Q.E.F. 


ELEMENTS  OF  PLANE  GEOMETRY.  113 

PROBLEM   VIII. 
222.   CH^ven  two  angles  of  a  triangle,  to  find  the  third  angle. 
Let  a  and  h  be  the  given  Z.8. 


./ 


A -CL-lw 


P 
Draw  the  indefinite  line  A  B. 

At  any  point  in  A  B,  as  P,  construct 

/-  m  =  Z-  a,  and  Z.  n  =  Z.  5,  as  in  (219). 

Then       Z.  c  is  the  third  Z.  of  the  A.         (77)     Q.  E.  F. 

PROBLEM    IX. 

223.  7\vo  sides  and  the  included  angle  of  a  triangle  being 
given,  to  construct  the  triangle. 

Let  m  and  n  be  the  given  sides,  and  a  the  included  Z. . 

E 


A_ 

.  A 

K 

m 

n 

\ 

Draw  the  indefinite  line  AB. 

C 

Cutoff 

AC  = 

=  n. 

At  A  construct  Z_  h  -- 

=  Z^a. 

On  AE  Q\xt  oS  AD  = 

m. 

Draw  CD. 

Then  A  CD  is  the  required  A.  (83)     Q.  E.  F. 

10* 


114  ELEMENTS  OF  PLANE  GEOMETRY. 


PROBLEM  X. 

224.  One  side  and  two  adjacent  angles  of  a  triangle  being 
given,  to  construct  the  triangle. 

Let  AB  he  the  given  side,  and  a  and  b  the  given  Z_g. 

\V 

/\ 

/  \ 


At  A  construct  Z.  c  =  Z^a,  and  at  B  construct  Z-  d  =  Z-b. 

The  sides  AD  and  BE  intersect  at  (7,  ' 

and  ABC  is  the  required  A.  (84)     Q.  E.  F. 

225.  Scholium.  This  problem  is  not  possible  if  the  sum 
of  the  two  given  angles  is  equal  to  or  greater  than  two  right 
angles. 


ELEMENTS  OF  PLANE  GEOMETRY.  115 


PROBLEM   XI. 

226.  Given  the  three  sides  of  a  triangle,  to   construct  the 
triangle. 


Let  m,  n,  and  o 

be  the  three  sides  of  a  A. 

n 

A 

m 

's. 

0 

\ 

Z 
A 

\ 

D         1 

Draw  an  indefinite  line  AB. 

It  off  AD  =  0. 

With  ^  as  a  centre  and  m  as  a  radius,  describe  an  arc; 
and  with  D  as  a  centre  and  n  as  a  radius,  describe  an  arc 
cutting  the  first  at  C. 

Drawee  and  DG 

Then  ADC  is  the  required  A.      (86)     Q.  E.  P. 

227.  Scholium. — This  problem  is  not  possible  if  one  side 
is  equal  to  or  greater  than  the  sum  of  the  other  two  sides. 


116 


ELEMENTS  OF  PLANE  GEOMETRY. 


PROBLEM   XII. 

228.  Gfiven  two  sides  of  a  triangle  and  the  angle  opposite 
one  of  them,  to  cdnstrud  the  triangle. 

Let  m  and   n   be   the  given  sides,  and  a  the  given  Z. 
opposite  the  shorter  side  m. 

I.  When  the  given  angle  is  acute,  and  the  side  opposite  is  less 
than  the  other  given  side. 


D 


n 

A. 

Xa 

y' 

::;*^^ 

.r 


...v ^ 


Construct  /^  h  =  /L  a,  and  on  AD  cut  o^ AC  =  n. 

With  (7  as  a  centre  and  m  as  a  radius,  describe  an  arc 
cutting  the  side  AE  at  B  and  P. 

Draw  CB  and  CP. 

Then  either  ABC  or  APC  is  the  required  A,  and   there 
are  two  solutions  to  the  problem. 

When  m  equals  the  J_  CF,  there  is  but  one  construction. 

When  m  is  less  than  CF,  the  problem  is  not  possible. 


ELEMENTS  OF  PLANE  GEOMETRY.  117 


II.  When  the  given  angle  is  acute,  right,  or  obtuse,  and  the 
side  opposite  is  greater  than  the  other  given  side. 


E 


m 

Cy^' 

n 

X 

Jv:</< ^,/._. 

F 

y^  d 

d\A                                              /B 

When  the  given  Z_  a  is  acute,  construct  Z^  b  =  Z^  a,  and 
cut  off  on  the  side  AE,  AC  =  n. 


With  (7  as  a  centre  and  m,  the  greater  side,  as  a  radius, 
describe  an  arc  cutting  the  side  AF  at  B,  and  AF  produced 
atZ). 

Draw  CB  and  CD. 

Then  ABC  is  the  required  A;  and  there  is  but  one 
solution.  Q.  E.  F. 

When  the  given  Z.  is  obtuse,  as  c,  the  A  A  CD  is  the 
required  one ;  and  there  is  but  one  solution. 

When  the  given  Z.  is  a  L,  the  problem  has  two  solutions. 
Let  the  pupil  give  the  construction. 

229.  Scholium. — The  problem  is  not  possible  if  the  given 
angle  is  right  or  obtuse,  and  the  side  opposite  is  less  than  the 
other  given  side. 


118  ELEMENTS  OF  PLANE  GEOMETRY.    . 


PROBLEM    Xin. 

230.  Given  two  sides  and  the  included  angle  of  a  parallelo- 
gram, to  construct  the  parallelogram. 

Let  m  and  n  be  the  given  sides,  and  a  the  included  Z. . 


Construct  /Lh  =  Z^  a,  and  on  the  sides  cut  off  J.  C  and 
AB  respectively  equal  to  m  and  n. 

With  ^  as  a  centre  and  J.C  as  a  radius,  describe  an  arc; 
with  C  as  a  centre  and  AB  as  a  radius,  describe  an  arc 
cutting  the  other  at  D. 

Draw  BD  and  CD. 

Then  ABCD  is  the  required  EJ.  (132)     Q.E.F. 


ELEMENTS  OF  PLANE  GEOMETRY.  119 


PROBLEM   XIV. 

231.  To  find  the  centre  of  a  given  drde, 
Jjet  ABC D  be  a  given  O- 


From  any  point  in  the  circumference,  as  B,  draw  two 
chords,  AB  and  BC. 

Bisect  AB  and  BC  hj  -Ls  as  in  (215). 

The  point  0,  the  intersection  of  the  JLb,  is  the  required 
centre.  (191)    Q.  E.  F. 


120 


ELEMENTS  OF  PLANE  GEOMETRY, 


PROBLEM   XV. 

232.  At  a  given  point  in  the  circumference  of  a  circle^  to 
draw  a  tangent  to  the  circle. 


Let  P  be  the  given  point  in  the  circumference  of  the  O 
EDP. 


If  the  centre  is  not  given,  find  it  by  (231). 


Draw  the  radius  OP,  and  at  P  draw  AB  _L  to  OP. 


Then  AB  is  the  required  tangent. 


(194)    Q.E.F. 


ELEMENTS  OF  PLANE  GEOMETRY.  121 


PROBLEM    XVI. 

233.   TJirough  a  given  point  without  a  given  circle,  to  draw  a 
tangent  to  the  circle. 

Let  0  be  the  centre  of  the  given  O ,  and  P  the  given  point. 


:MP 


Draw  OP,  and  upon  it  describe  a  circumference  cutting 
the  given  circumference  at  A  and  C. 

Draw  AP  and  CP;  also  the  radii  OA  and  OC. 

/_OAP=rL',  (208) 

AP  is  _L  to  OA,  and  is  tangent  to  the  O-  (194) 

Likewise  we  can  prove  that  CP  is  a  tangent.  Q.  E.  F. 

234.  Cor. — From  any  point  without  a  given  circle,  two  equal 
ta7igents  to  the  circle  can  be  drawn. 


11 


122 


ELEMENTS  OF  PLANE  GEOMETRY. 


PROBLEM    XVII. 

235.   To  inscribe  a  circle  in  a  given  triangle. 
Let  ABC  hQ  the  given  A. 


Bisect  i\\Q/-sABC  and  CAB. 

The  bisectors  meet  in  some  point,  as  0.  ' 

From  0  draw  OF.OD,  and  OE,  J- to  the  sides  of  the  A. 

The  J_3  are  all  equal.  (104) 

With  0  as  a  centre  and  OF  as  a  radius,  describe  a  circle. 

Then  FED  is  the  required  O.  (181)     Q.  E.  F. 


ELEMENTS  OF  PLANE  GEOMETRY. 


123 


PROBLEM    XVIII. 

236.  On  a  given  straight  line^  to  describe  a  segment  of  a 
circle  which  shall  coyitain  a  given  angle. 

Let  AB  be  the  given  line,  and  a  the  given  Z. . 
E 


/        /     \    N 

/F       \        \ 


A^- 


/^-. 


\     / 


0        \ 


D 


'^YB 


At  B  construct  the  Z.  AB  C  =^  /-  a. 

Draw(^^_Lto  jBCat^. 

Bisect  AB,  and  at  its  middle  point  erect  the  J_  BF. 

With  0,  the  intersection  of  i)i^  and  G^J5,  as  a  centre  and 
0^  as  radius,  describe  a  circumference. 

Now,  BC  is  J_  to  the  radius  OB; 

BC  is  R  tangent  to  the  circle.  (1^4) 

The  Z.ABC  is  measured  by  *  arc  AB.  (211) 

But  Z.  b,  the  Z.  inscribed  in  the  segment  J.  ^^,  is  measured 
by  i  arc  AB ;  (206) 


AEB  is  the  required  segment. 


Q.E.F. 


124  ELEMENTS  OE  PLANE  GEOMETRY. 


EXERCISES    1^    INYENTIOK 


THEOREMS. 

1.  In  any  circumscribed  quadrilateral,  the  sura  of  two 
opposite  sides  is  equal  to  the  sum  of  the  other  two  sides. 

2.  A  quadrilateral  is  inscriptible  if  two  of  its  opposite 
angles  are  supplements  of  each  other. 

3.  The  bisectors  of  the  angles  formed  by  producing  the 
opposite  sides  of  an  inscribed  quadrilateral  intersect  at  right 
angles. 

4.  If  a  circle  is  described  on  the  radius  of  another  circle, 
any  straight  line  drawn  from  the  point  of  contact  to  the 
outer  circumference  is  bisected  by  the  interior  one. 


PROBLEMS. 

1.  To  trisect  a  right  angle. 

2.  Given  two  lines  that  would  meet  if  sufficiently  pro- 
duced, to  find  the  bisector  of  their  included  angle  without 
finding  its  vertex. 

3.  To  draw  a  common  tangent  to  two  given  circles. 

4.  Inscribe  a  square  in  a  given  rhombus. 

5.  To  construct  a  square,  given  its  diagonal. 

6.  Construct  an  angle  of  30°,  one  of  60°,  one  of  120°,  one 
of  150°,  one  of  45°,  and  one  of  135°. 

7.  Construct  a  triangle,  given  the  base,  the  angle  opposite 
the  base,  and  the  medial  line  to  the  base. 


ELEMENTS  OF  PLANE  GEOMETRY.  125 

8.  Construct  a  triangle,  given  the  vertical  angle,  and  the 
radius  of  the  circumscribing  circle. 

9.  Construct  a  triangle,  given  the  base,  the  vertical  angle, 
and  the  perpendicular  from  the  extremity  of  the  base  to  the 
opposite  side. 

10.  Construct  a  triangle,  given  the  base,  an  angle  at  the 
base,  and  the  sum  or  difference  of  the  other  two  sides. 

11.  Construct  a  square,  given  the  sura  or  difference  of  its 
diagonal  and  side. 

12.  Describe  a  circle  cutting  the  sides  of  a  square,  so  as  to 
divide  the  circumference  at  the  points  of  intersection  into 
eight  equal  arcs. 

13.  Through  any  point  within  a  circle,  except  the  centre, 
to  draw  a  chord  which  shall  be  bisected  at  that  point. 


11* 


126  ELEMENTS  OF  PLANE  GEOMETRY. 


BOOK  IV. 

AEEA    A1^T>    RELATION    OF 
POLYGONS.      . 


DEFINITIONS. 

237.  Similar  Polygons  are  polygons  which  are  mutually 
equiangular,  and  have  their  homologous  sides  proportional. 

238.  The  Area  of  a  polygon  is  its  quantity  of  surface;  it 
is  expressed  by  the  number  of  times  the  polygon  contains 
some  other  area  taken  as  a  unit  of  measure.  The  unit  of 
measure  usually  assumed  is  a  square,  a  side  of  which  is  some 
linear  unit;  as,  a  square  inch,  a  square  foot,  etc. 

239.  Equivalent  Figures  are  such  as  have  equal  areas. 


ELEMENTS  OF  PLANE  GEOMETRY. 


127 


AREAS. 


THEOREM    I. 

240.  Tlie  area  of  a  rectangle  equals  the  product  of  its  base 
and  altitude. 

Let  ABCD  be  a  rectangle,  AB  the  base,  and  ^(7  the 
altitude. 


D 


B 


To  prove  that  the  area  of  ABCD  ^  AB  X  AC. 

Let  AE  be  a  common  unit  of  measure  of  the  sides  AB 
and  A  C,  and  suppose  it  to  be  contained  in  AB  5  times,  and 
in  J.C  3  times. 

Apply  AE  to  AB  and  AC,  dividmg  them  respectively  into 
five  and  three  equal  parts. 

Through  the  several  points  of  division  draw  -Ls  to  the 
sides. 


The  rectangle  will  then  be  divided  into  equal  squares,  as 
the  angles  are  all  Ls,  and  the  sides  all  equal.  (130) 


128  ELEMENTS  OF  PLANE  GEOMETRY. 


Now,  the  whole  number  of  these  squares  is  equal  to  the 
number  in  the  row  on  AB  multiplied  by  the  number  of 
rows,  or  the  number  of  linear  units  in  AB  multiplied  by  the 
number  in  J.C. 

Now,  this  is  true,  whatever  may  be  the  length  of  the 
common  unit  of  measure;  hence  it  is  true  if  it  is  infinitely 
small,  as  is  the  case  when  the  sides  are  incommensurable. 
Therefore,  in  any  case,  the  proposition  is  true.  Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


129 


THEOREM    II. 


241.  Rectangles  are  to  each  other  as  the  products  of  their 
bases  and  altitudes. 

Let  R  and  r  denote  the   areas  of  the   rectangles   whose 
bases  are  B  and  b,  and  whose  altitudes  are  A  and  a. 


To  prove  that  R  :  r   ::   A  X  B  :  a  X  b, 
R  =  AXB, 
and  r=  a  X  b;  (240) 

R  :  r  ::  A  X  B  :  a  X  b.      (159)      Q.E.D. 


242.  Cor. — Rectangles  having  equal  bases  are  to  each  other 
as  their  altitudes;  rectangles  haviiig  equal  altitudes  are  to  each 
other  as  their  bases. 


130  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    III. 

243.   The  area  of  a  parallelogram  equals  the  product  of  its 
base  and  altitude. 

Let  ABEC  be  a  ZZ7,  AB  its  base,  and  CD  its  altitude. 

0  E 


D  B 

To  prove  that  the  area  of  CD  ABEC  ==  AB  X  CD. 

Construct  the  rectangle  DFEC  having  the  same  base  and 
altitude  as  the  CJ. 

AC  =  BE,  and  DC  =FE; 
RAACD  =  IIA  BEF.  (91) 

Remove  A  A  CD,  and  rectangle  DFEC  remains. 
Remove  A  BEF,  and  O  ABEC  remains; 
rectangle  DFEC  =  CO  ABEC. 
But  the  area  of  the  rectangle  DFEC  =  AB  X  CD; 

the  area  of  the  o  ABEC  =  AB  X  CD.   Q.  E.  D. 

244.  Cor.  1. — Parallelograms  are  to  each  other  as  the  pro- 
ducts of  their  bases  and  altitudes. 

245.  Cor.  2. — Parallelograms  having  equal  bases  are  to  each 
other  as  their  altitudes;  parallelograms  having  equal  altitudes 
are  to  each  other  as  their  bases. 

246.  Cor.  3. — Parallelograms  having  equal  bases  and  equal 
altitudes  are  equivalent  figures. 


ELEMENTS  OF  PLANE  GEOMETRY.  131 


THEOREM   IV. 

247.   The  area  of  a  triangle  equals  half  the  product  of  its 
base  and  altitude. 

Let  ABC  be  a  A,  AB  its  base,  CD  its  altitude. 


B        ^ 

To  prove  that  the  area  of  the  A  ABC  =  *  {AB  X  CD). 

Through  C  draw  CE  II  to  AB,  and  through  A  draw  AE 
II  to  BC. 

The  A  ABC  'is  half  of  the  o  ABCE;  (129) 

but  the  area  of  the  CJ  ABCE  ==  AB  X  CD;  (243) 

the  area  of  the  A  ABC  =  i  (AB  X  CD).    Q.  E.  D. 

248.  Cor.  1. —  Triangles  are  to  each  other  as  the  products  of 
their  bases  and  altitudes. 

249.  Cor.  2. —  Triangles  having  equal  bases  are  to  each  other 
as  their  altitudes ;  triangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 

250.  Cor.  3. —  Triangles  having  equal  bases  and  equal  alii' 
tudes  are  equivalent  figures. 


132 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  V. 

251.   The    area  of  a  trapezoid  is  equal  to  half  the  sum  of 
its  parallel  sides  multiplied  by  its  altitude. 

Let  ABJJD  be  a  trapezoid,  AB  and  DC  its  II  sides,  and 
EF  its  altitude. 


^^-k 


"■^B 


To  prove  that  the  area  of  ABCD  =  ]  {AB  -^  DC)  EF, 

Draw  the  diagonal  BD,  forming  As  ABD  and  CDB. 

1(he  area  of  the  A  ABD  =  I  AB  X  EF, 

and  the  area  of  the  A  CDB  =1  DC  X  EF;       (247) 

.'.l^ABD^l\CDB=^2iVQ?.oiABCD=l{AB-\-DC)EF. 

Q.  E.  D. 

252.  CoR.--27ie  area  of  a  trapezoid  is  equal  to  the  product 
of  the  line  joining  the  middle  points  of  its  non-parallel  sides 
and  its  altitude. 

253.  Scholium. — The  area  of  an  irregular  polygon  is 
found  by  finding  the  areas  of  the  several  triangles  into  which 
it  can  be  divided.  In  survey- 
ing, the  method  usually  resorted 
to  is  to  draw  the  longest 
diagonal,  and  to  draw  perpen- 
diculars to  this  diagonal  from  the  other  vertices,  thus 
dividing  the  polygon  into  rectangles,  right-angled  triangles, 
and  trapezoids.  The  areas  of  these  figures  are  then  readily 
found. 


ELEMENTS  OF  PLANE  GEOMETRY. 


133 


SQUARES   O^  LIIsTES. 


THEOREM   VI. 

254.  Tlie  square  described  on  the  hypothenuse  of  a  right- 
angled  triangle  is  equivalent  to  the  sum  of  the  squares  on  the 
other  two  sides. 


Let  ABC  be  a  KA,  right-angled  at  C. 
G 


H 


\cy 


M 


To  prove  that  AB    =  AC   -]-  BG 


On  AB,  AC,  and  BC,  construct  the  squares  AE,  AG,  and 
BH. 

Through  C  draw'C^  II  to  BE,  and  draw  ^i^and  CD. 
/^  ACB  =  SiL  (Hyp.),  and  Z.  a  ::=^  a  L ;      (Cons.) 
GCB  is  a  straight  line. 

12 


134  ELEMENTS  OF  PLANE  GEOMETRY. 


For  a  like  reason  A  CH  is  a  straight  line, 

AB  =..  AD,  and  AF  =  A  C,  (Cons.) 

and    Z_  FAB  =  Z.  CAD,  each  being  a  L  ~\-  -^  b; 

.-.     .  A  J.J5i'^=  A  ^CD.  (83) 

Square  AG  and  A  ABF  have  a  common  base  J.i^  and  a 
common  altitude ; 

.  • .     Square  A  G  is  double  A  ABF;  (240)  and  (247) 

and  for  a  like  reason 

/U  AK  is  double  A  A  CD. 

But  A  A  CD  =  A  ABF; 

ZZ7  AK  =  square  A  G. 

Likewise  we  can  prove  ZZ7  BK  =  square  BH; 

.-.    cj  AK-^  OJ  BK=AB'' =  AZ''  X  BC\    (Q.E.D.) 

255.  Cor.  1. —  TJie  square  on  either  side  about  a  right-angled 
triangle  is  equivalent  to  the  square  on  the  hypothenuse  minus 
the  square  on  the  other  side. 

256.  Cor.  2. — The  square  on  the  diagonal  of  a  square  is 
double  the  given  square.  i 

257.  Cor.  Z.  —  The  diagonal  and  the  side  of  a  square  are 

incommensurable. 

Let  d  be  the  diagonal,  and  a  the  side  of  a 
square. 

Then  d^  =  a'  -\-  a' =^  2  a\ 

Extract  the  square  root  of  each  member. 

Then  d  —  a  \/'2. 

Divide  by  a; 

d  ,_ 

then  —  =  i/2  =  1. 41421 +. 


ELEMENTS  OF  PLANE  GEOMETRY.  135 


PROJECTIOK 


DEFINITIONS. 

258.  The  Projection  of  a  Point  upon  an  indefinite 
straight  line  is  the  foot  of  the  perpendicular  drawn  from 
the  point  to  the  line. 

Thus,  the  projection  of  the  point  C  upon  the  line  AB  is 
the  point  E. 

C 


E 


The  Projection,  of  a  Finite  Straiffht  Line  upon  an  in- 
definite one,  is  the  part  of  the  line  intercepted  between  the 
perpendiculars  drawn  from  the  extremities  of  the  finite  line. 
Thus,  EF  is  the  projection  of  CD  upon  AB. 


If  one  extremity  of  CD  rests  upon  the  other  line  AB,  then 
the  projection  of  CD  is  ED. 


136 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  VII. 

259.  In  any  triangle,  the  square  on  the  side  opposite  an  acute 
angle  equals  the  sum  of  the  squares  of  the  other  two  sides  minus 
twice  the  product  of  one  of  those  sides  and  the  projection  of  the 
other  upon  that  side. 

In  the  A  ABC,  let  c  be  an  acute  /_,  and  PC  the  pro- 
jection of  AC  upon  BC. 


To  prove  that  AB^  =r  BC'  +  ^C'  —  2  BC  X  PC. 

If  P  falls  on  the  base, 

PB  =  BC—Pa 
If  P  falls  on  the  base  produced, 

PB  =  PC—BC. 
In  either  case,  by  Algebra, 

PB'  :=  BC'  +  PC'  —  2BCX  PC, 

Add  AP  to  each  member; 
then  AP'  +  PB'  =  BC"  -\-  AP'  -\-PC'  —  2BCX  PC. 

But  AB'  =  AP'  +  PB\ 

and  AXf  ^AP'  -{-  PC'-  (254) 

Substitute  AB  and  AC  for  their  equals; 
then  AB'  =  BC'  -f  AC'  —  2BCX  PC.      Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


137 


THEOREM  VIII. 

260.  In  any  obtuse-angled  triangle,  the  square  on  the  side 
opposite  the  obtuse  angle  equals  the  sum  of  the  squares  of  the 
other  two  sides  plus  twice  the  product  of  one  of  those  sides 
and  the  projection  of  the  other  upon  that  side. 

In  the  A  ABC,  let  c  be  the  obtuse  Z-,  and  PC  the  pro- 
jection oi  AC  upon  BC  produced. 


To  prove  that  AB'  =-  B  C^  +  AC'  -\-  2  BC  X  PC. 

PB  =  BC  -\-  PC. 
By  Algebra,  PS  =  BC'  +  PC'  -\- 2  BC  X  PC. 
Add  AP^  to  each  member. 

Then  AP'  -f  PB'  =  BC'  +  Xp'  -j-  PC' -{- 2  B C  X  PC. 
But  AB'  =  AT'  +  PB\ 
and  AG''  =  A~P'  +  PC";  (254) 

Substitute  A  B^  and  A  C^  for  their  equals. 
Then  ArB"-  =  BC'  -f  ^'  -{-  2BCX  PC.  Q. E.  D. 

12* 


138 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    IX. 


261.  In  any  triangle,  if  a  medial  line  is  drawn  to  the  base: 

1.—  The  sum  of  the  squares  of  the  two  sides  equals  twice  the 
square  of  half  the  base  plus  twice  the  square  of  the  medial 
line. 

II. — Tlie  difference  of  the  squares  of  the  two  sides  equah 
twice  the  product  of  the  base  and  the  projection  of  the  medial 
line  upon  the  base. 

In  the  A  ABC,  let  AD  be  the  medial  line,  and  DP  the 
projection  of  AD  upon  the  base  BC. 


To  prove 

I.— Tliat  AB'  +  AC'  =  2  BD'  +  2  A^\ 
II.— That  AB"  —  AC'  =  2BCX  DP. 
If  AB  >  ^  C,  Z.  a  is  obtuse,  and  Z.  6  is  acute. 
Then  AB'  =  BD'  -f-  AD'  +  2  J5X>  X  DP,  (260) 

and  AC'  =  DC' ^  A~D'  —  2DCXDP.  (259) 

Add  these  equations,  observing  that  BD  =  DC. 
Then  jiB'  +  AC'  =  2  BD'  +  2  AJf- 
Subtract  the  second  equation  from  the  first. 
Then  AB'  —  AC'  =  2BCX  DP.  Q. E. D. 


ELEMENTS  OF  PLANE  GEOMETRY.  139 


THEOREM    X. 

262.  The  sum  of  the  squares  of  the  sides  of  ayiy  quadrilateral 
equals  the  sum  of  the  squares  of  the  diagonals  plus  four  times 
the  square  of  the  line  joining  the  middle  points  of  the  diagonals. 

In  the  quadrilateral  ABCD,  let  EF  be  the  line  joining 
the  middle  points  of  the  diagonals  JBD  and  AC. 
A 


D 

To  prove  that 

AB'  +  BG'  4-  CW  +  DA'  =  AC"  +  BD"  +  4  ET' 
Draw  DE  and  BE. 

Aff  -h  BC"  =  2  AE"  -f-  2  BE\ 
and  7fD'  -f  ^A'  =  2  AE'  +  2 ~DE''  (261) 

Add  these  equations. 


Then  AB'-^BC'-hCD'+ DA  =  4  AE'-i- 2  {BK-^DE') 
Now,  5:e'  +  :dI;'  =^2BF"  -{-2  lEF';  (261) 

But  4  1:e'  =  (2  AEf  =  A~C\ 

and  4  BF'  =  (2  jBjP)'  =  BD'; 

substituting,  JB'^  4.  ;g^2  ^  05'  +  ^^'  =  ^'  + 

^^^  4-  4  EF'\     Q.  E.  D. 

263.  Cor. — In  any  parallelogram,  the  sum  of  the  squares  of 
the  sides  equals  the  sum  of  the  squares  of  the  diagonals. 


140  ELEMENTS  OF  PLANE  GEOMETRY. 


PBOPOETIOJ^AL  LIll^ES. 


THEOREM  XI. 

264.  If  a  number  of  parallels  cutting  two  straight  lineSy 
intercept  equal  parts  on  one  of  the  lines,  they  also  intercept 
equal  parts  on  the  other. 

Let  the  lis  AB,  CD,  EF,  GH,  intersect  MN  and  OP, 
intercepting  on  MN  equal  parts  A  C,  GE,  EG. 

M    0 


To  prove  that  the  lis  intercept  on  OP  equal  parts  BD,  DF, 
FH. 

Through  the  points  B,D,  F,  draw  BQ,  BE,  FS,  II  to  3IN. 

AC=BQ,CE  =  DR,  and  EG  =-  FS.        (130) 

But       AC=  CE  =  EG ;  (Hyp.) 

BQ  =DB  =  FS. 

Now,  in  the  As  BQD,  DBF,  and  FSH, 

/L  a  =  Z.  c  =  Z.  e,  and  /.  b  =  Z.  d  =  A.  f;       (62) 

/\BQD=  A  DBF  =  A  FSH,  (84) 

and  BD  =  DF  =  FH.  Q.  E.  D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


141 


THEOREM    XII. 

265.  A  line  parallel  to  the  base  of  a  triangle   divides  the 
other  two  sides  proportionally. 

In  the  A  ABC,  let  the  line  DEhe  parallel  to  AB. 


To  prove  that  CA  :  CD  ::  CB  :  CE. 

Suppose  CA  and  CD  to  have  a  common  unit  of  measure, 
as  AG,  and  suppose  it  to  be  contained  in  CA  eight  times, 
and  in  CD  five  time. 

Then  CA   :  CD  ::   8  :  5. 

Draw  lis  to  AB  through  the  points  of  division  on  CA. 

Then  CB  is  divided  into  eight  equal  parts,  (264) 


(164) 


Now,  this  is  true,  whatever  may  be  the  length  of  the 
common  unit  of  measure ;  hence  it  is  true  if  it  is  infinitely- 
small,  as  is  the  case  when  the  lines  are  incommensurable. 
Hence  in  any  case  the  proposition  is  true.  Q.  E.  D. 


and 

CB  :  CE  : 

:   8  :  5. 

But 

CA  :  CD  : 

:   8  :  5; 

,  * . 

CA  :  CD  : 

:  CB  :   CE. 

142  ELEMENTS  OF  PLANE  GEOMETRY. 

266.  Cor.  1. — By  (163),  the  proportion  becomes 

CA  —  CD  I  CA  ::  CB  —  CE  :  CB, 
or  DA  :  CA   ::   EB  :    CB. 

267.  Cor.  2. — By  (161),  the  last  proportion  gives 

CA  :  DA   ::  CB  :  EB,  which  by  (163) 

gives       CA  —  DA  :  DA   ::  CB  —  EB  :  EB, 
or  CD  :  DA   ::   CE  :  EB; 

CD  :   CE  ::  DA  :  EB.  (160) 

THEOREM    XIII. 

268.  J.  straight  line  which  divides  two  sides  of  a  triangle 
proportionally  is  parallel  to  the  third  side.   . 

In   the  A  ABC,  let   DE  divide   the  sides  AB  and  AC 
proportionally. 


To  prove  that  DE  is  11  to  BC. 

Suppose  DE  to  be  drawn  II  to  BC. 

Then  AB  :  AD  ::  AC  :  AF.  (265) 

But    AB  :  AD  ::  AC  :  AE;  (Hyp.) 

AC  :  AF  ::  AC  :  AE;  (164) 

AF  =  AE,  which  cannot  be  unless  DF  coincides  with 
DE; 

DE  13  W  to  BC,  Q.E.D, 


ELEMENTS  OF  PLANE  GEOMETRY. 


143 


SIMILAEITY  OF  PO 


THEOREM  XIV 

269.   Two  triangles  mutually  equiangular  are  'si' 

Let  the  Z_s  a,  6,  c,  of  the  A  ABC  be  respectively  equal 
to  the  Z-sd,  e,  /,  of  the  A  DEF. 


To  prove  that  As  ABC  and  DEF  are  sit 

Place  the  A  ABC  on  A  DEF  so  as  to  make  Z_  a  coin- 
cide with  its  equal  Z_  d. 

Then  A  ABC  takes  the  position  DGH. 

Since  A  5  =  Z.  m  =-  /.  ^  G^IT  is  II  to  EF ;  (64) 

Z>^  :  ZX^  ::   DF  :  DH, 

or  DE  :   ^5  ::   7)i^  :  AC.  (265) 

Likewise  we  can  prove  that 

DE  :  ^^  ::   EF  :  J?C. 

the  A  a  are  similar.        (237)     Q.E.D. 

270.  Cor. —  Two  triangles  are  similar  if  two  angles  of  the 
one  are  respectively  equal  to  two  angles  of  the  other ;  two  right- 
angled  triangles  are  similar  if  one  has  an  acute  angle  equal 
to  an  acute  angle  of  the  other. 


144 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    XV. 

271.   Two  triangles  whose  homologous  sides  are  proportional 
are  similar. 

In  the  As  ABC  and  DEE  let 

DE  :  AB  ::   DF  :  AC  ::   EF  :  BC. 


To  prove  that  As  ABC  and  DEE  are  similar. 

On  DE  and  DF,  cut  off  DG  and  DH  respectively  equal 
to  AB  and  A  C,  and  draw  GH. 

Since  DG  ==  AB,  and  DIT  :=  ^ C, 


DE  :  DG  ::   i)i^  :  DH; 

(Hyp.) 

GH  is  W  to  £i^: 

(268) 

Z-  a  =  Z.  b,  and  Z-  c  =  /-  d; 

(62) 

As  DG^IT  and  i).Ei^  are  similar, 

(269) 

and                     DE  :  DG  ::  J^Ji^  :   GH; 

(237) 

also                    DE  :  AB   ::   ^F  :  BC. 

(Hyp.) 

Of  these  two  proportions  take  the  forms 

DE      EF       ^  DE       EF 
DG      GH'^'''^  AB~  BC 

Divide;  then                  DG~  GH 

ELEMENTS  OF  PLANE  GEOMETRY. 


145 


But  DG  =  AB; 

GH=BC; 
A  DGH=-A  ABC.  (86) 

But  As  DGH  and  DEF  are  similar; 

As  ABC  and  DEF  are  similar.         Q.  E.  D. 


THEOREM  XVI. 

272.   Two  triangles  having  an  angle  in  each  equal  and  the 
including  sides  proportional,  are  similar. 

In  the  As^^C  and  DEF,  let 

L.  a=  Z-h,  and  DE  :  AB  ::   DF  :  AC. 

D 

A 


To  prove  that  As  ABC  and  DEF  are  similar. 

Place  A  ABC ox\  A  DEF  so  that  Z.  a  coincides  with  Z_  h. 

B  falls  somewhere  on  DE,  as  at  G,  and  C  falls  somewhere 
on  DF,  as  at  H. 

Then                DE  :  DG   ::   DF  :  DH;  (Hyp.) 

GH  i^W  to  EF.  (265) 

A  c  =  21  cZ,  and  Z_  e  =  Z.  /;  (62) 

As  DG^IT  and  DEF  are  similar.  (269) 

But  A  DGH  =  A  ABC; 

As  ^^  O  and  D^i^  are  similar. .         Q.  E.  D. 
13 


146  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XVII. 

273.  Two  triangles  having  their  sides  respectively  parallel  are 
similar. 

In  the  As  ^^C  and  DEF,   let   AB,  AC,  and  BC  hQ 
respectively  II  to  DE,  DF,  and  EF. 


To  prove  that  ^s  ABC  and  DEF  are  similar. 

Since  the  sides  are  II,  the  corresponding  Z_s  are   either 
equal  or  supplements  of  each  other.  (68)  and  (69) 

three  suppositions  can  be  made : 

(1)  Z.a4-Z.c?=2U  Z.^>  +  Z.e  =  2U  Z-c+Z./=2Ls. 

(2)  Z.a=  Z.d,      Z.  6  +  Z.  e  =  2  L?,  Z.  c  +  Z_/=  2  Ls. 

(3)Z.a=Z.(Z,  l.b=A-e;       .'.Z.c=Z.f. 

Now,  the  sum  of  all  the  Z_8  of  two  ^s  cannot  exceed  4  Ls; 

(76) 

the  third  supposition  is  the  only  one  admissible; 
AbABC  and  DEF  are  similar.         (269)     Q.  E.  P. 


ELEMENTS  OF  PLANE  GEOMETRY. 


147 


THEOREM   XVIII. 

274.   Tivo  triangles   having  their  sides   respectively  perpen- 
dicular to  each  other  are  similar. 

In  the  As  ABC  and  DEF,  let  the  sides  AB,  AC,  and 
^0  be  respectively  J-  to  DF,  FE,  and  DE. 


4  Ls. 
But 


To  prove  that  As  ABC  and  DEF  are  similar. 

Prolong  the  sides  of  the  A  DEF  till  they  meet  the  sides 
of  the  A  ABC 

The  sum  of  the  Z.s  of  the  quadrilateral  AGFH  equals 

(115) 
Z.S  d  and  e  are  Ls.  (Hyp.) 

Z.  a  +  Z.  c  =  2  Ls. 
But  Z.b  +  Z.c  =  2Ls,  (44) 

Z.b  =  /-a. 
Likewise  we  can  prove 

/L  0  =  Z-  m,  and  Z-  n  =  Z.  p. 
.'.     the  As  are  similar.  (269)     Q.  E.  D. 

275.  Scholium. — In  two  triangles  whose  sides  are  re- 
spectively parallel,  or  perpendicular,  the  homologous  sides 
are  the  parallel  sides,  or  the  perpendicular  sides. 


148  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    XIX. 

276.  Straight  lines  drawn  from  the  vertex  of  a  triangle  to  the 
hose  divide  the  base  and  its  parallel  proportionally. 

Let  CH  and  CK  be  straight  lines  drawn  from  the  vertex 
C  to  the  base  AB  of  the  A  ABC,  and  let  DE  he  parallel 
to  the  base. 


To  prove  that 

AH  :  DF  ::  HK  :  FG  ::  KB  :   GE. 

As  CAH,  CHK,  and  CKB  are  respectively  similar  to  As 
CDF,  CFG,  and  CGE.  (269) 

.-.  AH  :  DF  ::  CH  :  CF  ::  HK  :  FG  ::  CK  :  CG 

::  KB  :  GE.    (237)    Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


149 


THEOREM   XX. 

277.  Two  polygons  are  similar  if  they  are  composed  of  the 
same  number  of  triangles  similar  each  to  each  and  similarly 
placed. 

Let  the  As  ABE,  EBD,  and  DEC,  of  which  the  polygon 
ABCDE  is  composed,  be  respectively  similar  to  the  As 
FGM,  MGK,  and  KGH,  of  which  the  polygon  FGHKM 
is  composed. 


A  B  F  G 

To   prove    that    polygons  ABCDE    and  FGHKM   are 
similar. 

Z_  a  =  Z^  7n,  and  l^  h  =  /-  n.  (237) 

/^  c  =^  /-  0, 

and  /Ld=  Z^p.  ^  (237) 

Add  these  two  equations. 

Then  l.c-{-Z-d=^/Lo-]r'^P, 

or  Z.  AED  =  Z.  F3IK. 

Likewise  we  can  prove  Z_  EDO  =  Z_  MKH, 
and  Z^  ABC  =  A.  FGH; 

.  • .     the  polygons  are  mutually  equiangular. 

Also  AB  :  FG  ::  AE  :  FM  ::   EB  :  MG*  ::  ED  : 

3fK,  etc.;    (237) 

.  • .     the  polygons  have  their  homologous  sides  proportional. 

.-.     polygons  ABCDE  and  FGHKM  are  similar.  (237) 

Q.E.D. 

13* 


150 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    XXI. 

278.  Two  similar  polygons  can  he  divided  into  the  same 
number  of  triangles  similar  each  to  each  and  similarly  placed. 

Let  the  polygon  ABODE  be  similar  to  the  polygon 
FGHKMj  and  let  diagonals  be  drawn  from  the  vertices 
B  and  G, 


To  prove  that  the  As  ABE,  EBD,  and   DBG  are    re- 
spectively similar  to  the  As  FGM,  MGK,  and  KGH. 


(237) 
(272) 
(237) 


A  a  =  Z.  fZ,  and  ^^  :  FG  ::   AE  :  FM; 

^  ABC  i^  similar  to  A  FGM. 

Also  /-  h  -\-  /-  c  =  /-  e  -{-  /-  m. 

But  /.  b  =  /-  e; 

Z-  c  =  /-  m. 

And  EB  :  MG  ::   AE  :  FM  ::   ED  :  MK;       (237) 

A  EBD  is  similar  to  A  3IGK.  (272) 

Likewise  we  can  prove  that  A  DBG  ia  similar  to  A  KGH. 

Q.  E.  D. 

279.  Cor. —  Two  similar  polygons  can  be  divided  into  the 
same  number  of  triangles  similar  each  to  each  and  similarly 
placed,  by  drawing  lines  to  their  vertices  from  any  two  homolo- 
gous points. 


ELEMENTS  OF  PLANE  GEOMETRY. 


151 


THEOREM  XXII. 

280.   The  perimeters  of  two  similar  polygons  are  to  each  other 
as  any  two  homologous  sides. 

Let   the   polygon   ABODE    be   similar   to   the   polygon 
FGHKM,  and  let  p  and  P  denote  their  perimeters. 


To  prove  that  p  :  P  ::   AB  :  FG. 

AB  :  FG  ::   BC  :   GH  ::    CD  :  HK,  etc.;  (237) 

'.  AB-\-  BC-\-  DC,  etc.  :  FG -{- GH -\-  HK,  etc.  :: 

AB  :  FG,     (168) 


or 


p  :  P  ::  AB  :  FG. 


Q.E.D. 


281.  Cor. —  The  perimeters  of  two  similar  polygons  are  to 
each  other  as  any  two  homologous  lines. 


152  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    XXHI. 

282.  If  a  perpendicular  is   drawn  from  the  vertex  of  the 
right  angle  of  a  right-angled  triangle  to  the  hypotheniise  : 

I.  The  tivo  triangles  formed  are  similar  to  the  given  triangle 
and  similar  to  each  other. 

II.  The  perpendicular  is  a  mean  proportional  between  the 
segments  of  the  hypothenuse. 

III.  Each  side  of  the  right-angled  triangle  is  a  mean  pro- 
portional between  the  hypothenuse  and  the  adjacent  segment. 

Let  ABC  be  a  RA,  right-angled  at  C,  and  let  CD  be  the 
_L  to  the  hypothenuse  AB. 


I.  To   prove  that   the   /Sb  ACD,   ABC,   and    CBD   are 
similar  to  each  other. 

Z_  a  is  common  to  the  RAs  A  CD  and  ABC; 

A  ^07)  and  A  ^^C  are  similar.  (270) 

For   a   like   reason  As  CBD  and  ABC  are  similar,  and 
hence  similar  to  A  CD. 

II.  To  prove  that  AD  :  CD  ::  CD  :  DB. 

The  As  ACD  and  CBD  are  similar; 

AD  :  CD  ::  CD  :  DB.  (237) 


ELEMENTS  OF  PLANE  GEOMETRY.  153 


III.  To  prove  that  AB  '.AC  ::   AC  :  AD, 
and  that  AB  \  BC  ::   BC  :   BD. 

The  As  ACD  and  ABC  are  similar; 

AB  :  AC  ::  AC  :  AD,  (237) 

And  the  As  CBD  and  ABC  are  similar; 

AB  :  BC  ::   BC  :  BD.      (237)    Q.E.D. 

283.  Cor.  1. —  The  squares  of  the  sides  about  the  right  angle 
are  to  each  other  as  the  adjacent  segments  of  the  hypothenuse. 

From  the  last  two  proportions  we  have 

AC''  =  ABX  AD, 

and  BC'  =  AB  X  BD.  (156) 

Divide,  and  cancel  the  common  factor  AB. 

AC'       AD 


Then 


or 


BC       BD 


AC'  :  BC'  ::  AD  :  BD. 


154  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM  XXIV. 

284.    Two  triangles  having  an  angle  in  each  the  same  are  to 
each  other  as  the  products  of  the  sides  including  the  equal  angles. 

In  the  ^sABC  and  DEC  let  the  angle  c  be  common. 

C 


To  prove  that  £^  ABC  :  ^DEC  ::  CAxCB:  CDy^  CE. 

Draw  the  line  AE. 

A  ABC  :  A  AEC  ::  CB  :   CE, 

and  A  AEC  :   A  DEC  ::    CA  :  CD,  (249) 

Take  the  forms 

A  ABC  _  CB 
A  AEC~~  CE' 


and 


A  AEC        CA 


A  DEC ~  CD 
Multiply,  and  cancel  the  common  factor. 

A  ABC  _  CA  X  CB 
^^  A  DEC  ~~  CDX  CE' 

or   A  ABC  :  A  DEC  ::  CA  X  CB  :  CD  X  CE.   Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY.  155 


THEOREM  XXV. 

285.  If  two  chords  of  a  circle  intersect,  their  segments  are 
reciprocally  proportional. 

Let  the  chords  AB  and  CD  intersect  at  0. 


To  prove  that  CO  :  AO  ::  BO  :  DO. 
Draw  AD  and  CB. 

Z.  c  =  Z-  a,  and  Z.  6  =  Z.  c?;  (206) 

As  005  and  ilOZ>  are  similar,  (269) 

and  *     CO  :  AO  ::  BO  :  DO.    (237)    Q.E.D. 


156  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    XXVI. 

286.  If  two  secants  are  drawn  from  a  point  without  a  circle, 
the  secants  and  their  external  segments  are  reciprocally  j)ro- 
portional. 

Let  the  secants  PA  and  PB  be  drawn  from  the  point  P 
without  the  O. 


To  prove  that  PA  :  PB  ::  PE  :  PD. 

Draw  AE  and  BD. 

Z.  a=  Z-  b,  (206) 

and  /^  o  =  Z-  o; 

/\hPAE  and  PBD  are  similar,  (270) 

and  PA  :  PB  ::   PE  :  PD.        (237)      Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY.  157 


THEOREM    XXVII. 

287.  If  a  ta7igent  and  a  secant  are  drawn  from  a  point 
without  a  circle^  the  tangent  is  a  mean  proportional  between 
the  secant  and  its  external  segment. 

Let  the  tangent  FB  and  the  secant  PA  be  drawn  from 
a  point  P  without  the  O- 


To  prove  that  PA  :  PB  ::  PB  :  PC. 
Draw  AB  and  BC. 

Z_  a  is  measured  by  2  arc  CB,  (206) 

and  Z.  6  is  measured  by  ^  arc  CB;  (211) 

jL  a=  /:.  b. 
Also  Z.  G  =  Z-  G. 

As  PAB  and  PBC  are  similar,  (270) 

and  PA  :  PB  ::   PB  :  PC.      (237)     Q.  E.D. 


158 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM   XXVIII. 

288.  In  any  triangle,  the  product  of  two  sides  equals  the 
product  of  the  diameter  of  the  circumscribed  circle  and  the 
perpendicular  drawn  to  the  third  side  from  the  vertex  of  the 
opposite  angle. 

Let  a  O  be  circumscribed  about  the  A  ABCy  and  let  CD 
be  a  diameter  and  CE  a  _L  to  AB. 


To  prove  that  AC  X  CB  =  CDX  CE. 

Draw  DB. 

l.a=  JLd;  (206) 

RAs  AEC  and  CBB  are  similar,  (270) 

and                       AC  \    CD  w    CE  \   CB;  (237) 

AC  XCB  =  CD  X  CE.     (156)  Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY.  159 


THEOREM   XXIX. 

289.  In  any  triangle  the  product  of  two  sides  equals  the 
product  of  the  segments  of  the  third  side  formed  by  the  bisector 
of  the  opposite  angle  plus  the  square  of  the  bisector. 

Let  a  O  be  circumscribed  about  the  A  ABC,  and  let  the 
Z_  opposite  AB  be  bisected  by  CD. 

C 


E 

To  prove  that  CB  X  CA  =  BD  X  DA  -\-  CD\ 

Produce  CD  to  E,  a  point  in  the  circumference,  and  draw 
AE. 

l.b=  /-  e,  (206) 

and                                   Z_  c=  Z.  d;  (Cons.) 

As  GBD  and  CAE  are  similar,  (270) 

and                      CB  '.    CE  ::    CD  :    CA;  (237) 

CBX  CA  =  CEXCD.  (156) 

But  CEXCD  =  {DE-\-  CD)  CD=:DExCD-{-  CD'; 

CB  X  CA  =  DE  X  CD  -\-  CD\ 

Now,         DEXCD  =  BDX  DA.  (285) 

Substitute  BD  X  DA  for  its  equal  DE  X  CD. 

Then           CB  X  CA  =  DB  X  DA  ^  CD\  Q.  E.  D. 


160 


ELEMENTS  OF  PLANE  GEOMETRY. 


EELATIOISr    OF    POLTGOISrS. 


THEOREM     XXX. 

290.  Similar  triangles  are  to  each  other  as  the  squares  of  their 
homologous  sides. 

Let  the  As  ABC  and  EFG  be  similar. 


To  prove  that  AABC:   A  EFG   ::   AS"  :   EF\ 

Draw  the  altitudes  AD  and  EH. 

Then  A  ABC  :  A  EFG   ::  BC  X  AD  ::  FG  X  EH. 

(248) 

But  BC  :  FG^  ::  AB  :  EF, 

and  AD  :  ^il  ::   ^5  :  ^F;  (237) 

BCxADiFGxFH:  AB'  :  EF'        (169) 

Compare  this  with  the  first  proportion; 
then     A^^C:  A  EFG  ::  TB'  :  'EF\     (164)    Q.E.D. 


ELEMENTS  OF  PLANE  GEOMETRY. 


161 


THEOREM    XXXI. 

291.  Similar  polygons  are  to  each  other  as  the  squares  of 
their  homologous  sides. 

Let  the  polygons  ABODE  and  FGHKM  be  similar,  and 
denote  their  surfaces  by  s  and  S. 


To  prove  that  s  :  S  ::  AB'  :  FG\ 

Draw  the  diagonals  A  C,  EC,  and  FH,  MIT,  dividing  the 
polygons  into  homologous  As.  (278) 

A  ABC:  AFGH  ::  AC' 


Then  A  ABC  :  A  FGH  ::  AC  :  FH', 

and  A  ^ C^  :  A  FHM  : :  A&  :  FH)  (290) 

AABC:  AFGH  ::  AACE:  AFHM.       (164) 

Likewise  we  get  A  ^  C^  :  A  FII3f  : :  A  ECD  :  A  MHK; 

.-.    A  ABC  ^  A  ACE  -]-  A  ECD  :  AFGH+  A  FHM 

+  A  MHK  ::  A  ABC  :  A  FGH    (168) 

or  s  :  S  ::  A  ABC  :  A  FGH 

But  A  ABC  :  A  FGH  ::   AB'  :  FG';  (290) 

s  :  S  ::  AB'  :  FG\        (164)        Q.E.D. 

292.  Cor. — Similar  polygons  are  to  each  other  as  the  squares 
of  any  of  their  homologous  lines. 

14* 


162 


ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    XXXII. 

293.  If  similar  polygons  are  constructed  on  the  three  sides 
of  a  right-angled  triangle,  the  polygon  on  the  hypothenuse  is 
equivalent  to  the  sum  of  the  polygons  on  the  other  two  sides. 

Let  0,  P,  and  Q  be  the  similar  polygons  constructed 
respectively  on  AB,  BC,  and  AC,  the  three  sides  of  a  RA. 


To  prove  that  Q  =  0  -\-  P. 

0  :  P  ::  IB'  :  BC\ 
and  Q  :  P  ;:  AC'  :  BC\ 

From  the  first  proportion  we  get 

0  +  P  :  P  ::  IB'  -j-  BC"  :  BC" 
But  AB'  +  BC'  =  AC'; 

0  -\-  P  :  P  ::  AC'  :  BC\ 


(291) 
(162) 


ELEMENTS  OF  PLANE  GEOMETRY.  163 

Comparing  this  with  the  second  proportion,  we  get 
Q  '.  P  ::    0  -j-  F  :  P. 

Take  the  form  ^  =  — p — ,  and  multiply  each  member 
by  P. 
Then  Q  =  0  +  P.  Q.E.D. 


164  ELEMENTS  OF  PLANE  GEOMETRY. 


PKOBLEMS  IIT   CONSTRUCTION 


PROBLEM   I. 

294.   To  cut  a  given  straight  line  into  any  7iumber  of  equal 
parts. 

Let  AB  be  the  given  straight  line. 

^-=-::n r- ] r 1 iB 


c--~-^J      I      !      I 


■-J 


i  ! 


->.o 


From  one  extremity  A  draw  the  indefinite^  straight  line  A  0. 

Take  any  convenient  line,  as  A  C,  and  cut  it  off  on  A  0  as 
many  times  as  ^^  is  to  be  cut  into  equal  parts. 

Join  the  last  point  of  division,  as  P,  and  the  extremity  B 
of  the  given  line. 

Through  all  the  other  points  of  division  on  AO,  draw  lis  to 
PB. 

Then  AB  is  cut  into  equal  parts.  (264)     Q.  E.  F. 


ELEMENTS  OF  PLANE  GEOMETRY.  165 


PROBLEM   11. 

295.   To  cut  a  given  straight  line  into  parts  proportional  to 
given  straight  lines. 

Let  ABj  0,  p,  and  q  be  the  given  straight  lines. 


E-^O 


From  A  draw  the  indefinite  straight  line  A  0. 

Cut  off -40  =  0,  CD=p,  and  DE  =  q. 

Join  EB,  and  draw  DG  and  CF  II  to  EB. 

Then  the  parts  AF,  FG,   and  GB    are  proportional  to 
0,^,  and^.  (265)     '   Q.E.R 


166  ELEMENTS  OF  PLANE  GEOMETRY. 


PROBLEM   III. 


296.  To  construct  a  fourth  proportional  to  three  given  straight 
lines. 


Let  n,  0,  and  p  be  the  three  given  lines. 


U 

~ 

F 

n 

0 

i:-^ 

B 

_£. 

^^-a 

Construct  any  convenient  Z. ,  producing  the  sides  AB  and 
A  G  indefinitely. 

Cut  off'^D  =  o,AE  =  n,  and  EG  ==  p. 

Join  ED,  and  draw  GF  II  to  ED. 

Then  AE  :  AD  ::   EG    :    DF,  (267) 

or  n  :   0   ::  p  \  DF; 

DF  is  the  required  line.  Q.  E.  F. 

297.  Scholium.— If  EG^  is  equal  to  AD,  DF  \s  the  third 
proportional  to  AE  and  AD,  or  to  n  and  o. 


ELEMENTS  OF  PLANE  GEOMETRY. 


167 


PROBLEM    IV. 

298.  Given  two  straight  lines,  to  construct  a  mean  proportional 
between  them. 

Let  71  and  o  be  the  two  given  lines. 


or 


Draw  the  indefinite  straight  line  AE. 

Cut  off  AD  =  n,  and  DB  =  o. 

Describe  a  semi-circumference  on  AB,  and  at  D  erect  the 
-DC. 

Draw  .4  C  and  BG. 

Z.  AGB  :=  a  L;  (208) 

AD  :  DC  ::  DC  :  DB,  (282) 

n  :  DC  ::   DC  :  o; 

DC  13  the  required  line.  Q.  E.F. 


DEFINITION. 


299.  A  straight  line  is  said  to  be  divided  in  extreme  and 
mean  ratio,  when  the  greater  part  is  a  mean  proportional 
between  the  whole  line  and  the  less  part. 


168 


ELEMENTS  OF  PLANE  GEOMETRY. 


PROBLEM   V. 

300.   To  divide  a  given  straight  line  in  extreme  and  mean 
ratio. 

Let  AB  be  the  given  straight  line. 


/ 


i)\.''' 


O,-'- 


:\E 


At  the  extremity  B  erect  the 
AB. 


BC  equal  to  half  of 


With  C  as  a  centre  and  CB  as  a  radius,  describe  a  O . 

Draw  J.  C,  and  produce  it  till  it  meets  the  circumference  atE. 

On  AB  cut  off  AP  =  AD. 

Now,  AB,  being  _L  to  the  radius  CB  at  B,  is  a  tangent ; 

AE  :  AB  ::  AB  :  AD;  (287) 

AE  —  AB  :  AB   ::  AB  —  AD  :  AD.      (163) 

But         DE  =  2  CB  =  AB; 

AE  —  AB  =  AD  =  AP. 

AB  —  AD  =  AB  —  AP  =  PB. 

Substitute  AP  and  PB  for  their  equals. 

Then         AP  :  AB  ::   PB  :  ^P,  which  gives 

AB  :  AP  ::   AP  :  PB;  (161) 

.     AB  is  divided  in  extreme  and  mean  ratio  at  P.     (299) 

Q.E.F. 


ELEMENTS  OF  PLANE  GEOMETRY.  169 


PROBLEM   VI. 

301.  On  a  given  straight  line  to  construct  a  polygon  similar 
to  a  given  polygon. 

Let  MN   be   the   given   line,   and   ABCDE  the   given 
polygon. 


Q< 


\0 


M  N 


Divide  the  polygon  into  As. 

At  M  construct  Z.  m  --  Z_  a,  and  at  N  construct  /-  n 
=  Z.  6. 

Then  As  MNO  and  ABC  are  similar.  (270) 

Likewise   construct  As  MOP  and  MPQ  similar  to  As 
ACD  and  ADE. 

Then  polygons  MNOPQ  and  ABCDE  are  similar;  (277) 

MNOPQ  is  the  polygon  required.         Q.  E.  F. 


15 


170  ELEMENTS  OF  PLANE  GEOMETRY. 

PROBLEM  VII. 

302.   To  construct  a  triangle  equivalent  to  a  given  polygon. 
Let  ABODE  be  the  given  polygon. 


G 


''A 

/ 

/ 

^  / 

\ 

/' 

\  \ 

// 

\/ 

\ 

// 

— -1  F 

E 


Draw  the  diagonal   CAy  produce  EA,  draw  BG  W  to  CAj 
and  draw  CG. 

Then  As  BAG  and  GAG  have  a  common  base  AG  and 
a  common  altitude; 

.-.     .  ^BAG  =  ^GAG;  (250) 

polygon  GGDE  =  polygon  ABGDE. 

Draw  the  diagonal  GE,  produce  AE,  draw  DF  II  to  CEy 
and  draw  GF, 

Then  A  FGE  =  ADGE;  (250) 

.-.     ^  GGF  =  polygon  GGDE  =  polygon  ABGDE; 

GGF  is  the  required  A.  Q.  E.  F, 


ELEMENTS  OF  PLANE  GEOMETRY. 


171 


PROBLEM  VIII. 

303.   To  construct  a  square  equivalent  to  a  given  triangle. 
Let  ABC  be  the  given  A,  6  its  base,  and  a  its  altitude. 


Find  the  mean  proportional  x  between  a  and  one-half  of  h, 
by  (298). 


On  X  construct  the  square  S. 
Then  a:'  =  ^^  X  a  =  A  ABQ; 

S  is  the  required  square. 


(247) 
Q.  E.  F. 


304.  Scholium. — By  means  of  this  problem  and  the  pre- 
ceding one,  a  square  can  be  constructed  equivalent  to  any 
given  polygon. 


172 


ELEMENTS  OF  PLANE  GEOMETRY. 


PROBLEM    IX. 

305.   To  construct  a  square  equivalent  to  any  number  of  given 
squares. 

Let  m,  n,  o,  p,  and  q  be  the  sides  of  the  given  squares. 

F 


/     \      \ 

s 

m 

1 

1 

Of.        \  \ 

n 

1          "^N                            \             \ 

0 

!       ""v   \  \ 

A 1 :::^ 

P 
5 

1 

1 

Draw  AB  =  m. 
At  A  draw  AC  =  n  and 
At  C  draw  CD  ==  o  and 
At  D  draw  DE  =  p  and 
At  E  draw  EF  =  q  and 


to  AB,  and  draw  BC 
to  BC,  and  draw  BD. 
to  jBZ),  and  draw  BE. 
to  ^5,  and  draw  BF. 


On  JBi^  construct  the  square  S. 
Now,  BF'  =  EF'  -{-  BE'' 

=  EF'  -^  DE"  +  BD" 

=  EF'  -^  DE'  -{-  CD'  -i-  CB' 

=  EF'-\-  DE'  +  ~CD'  -\-  AXf  +  AF,       (254) 
or  S=  m^  +  w^  +  o'  4-  j9^  +  5^• 

/S  is  the  required  square.  Q.  E.  F. 

306.  Scholium  1. — By  means  of  this  and  the  two  pre- 
ceding problems,  a  square  can  be  constructed  equivalent  to 
the  sum  of  any  number  of  given  polygons. 

307.  Scholium  2. — If  m,  n,  o,  p,  and  q  are  homologous 
sides  of  given  similar  polygons,  BF  is  the  homologous  side 
of  a  similar  polygon  which  is  equivalent  to  the  sum  of  the 
given  polygons  (293). 


ELEMENTS  OF  PLANE  GEOMETRY. 


173 


PROBLEM   X. 

308.   To  construct  a  square  equivalent  to  the  difference  of  two 
given  squares. 

Let  0  be   the  side   of  the  larger  square,  and  p  the  side 
of  the  smaller. 


C 


M 


Construct  a  L  a,  and  cut  AB  =  p. 

With   5  as  a  centre   and  o  as  a  radius,  describe  an  arc 
cutting  AM  at  C. 

On  A  C  construct  the  square  S. 


Now, 


BC"  —  AB'  =  AC' 


S, 


or 


S  is  the  required  square. 


(255) 
Q.  E.  F. 


309.  Scholium  1. — By  means  of  this  problem  and  the  two 
immediately  preceding  (305),  a  square  can  be  constructed 
equivalent  to  the  difference  of  any  two  polygons. 

310.  Scholium  2. — If  o  and  p  are  the  homologous  sides 
of  two  given  similar  polygons,  AC  is  the  homologous  side 
of  a  similar  polygon  equivalent  to  the  difference  of  the  two 
given  polygons. 

15* 


174 


ELEMENTS  OF  PLANE  GEOMETRY. 


PROBLEM    XI. 

311.  To  construct  a  rectangle,  given  its  area  and  the  sum  of 
the  base  and  altitude. 

Let  AB  be  equal  to  the  given  sum  of  the  base  and 
altitude,  and  let  the  given  area  equal  that  of  the  square 
whose  side  is  a. 


Ell— 


■4- 


F  \ 


'B 


On  AB  as  a  diameter,  describe  a  semicircle,  and  at  A  erect 

the_L^C  =  a. 

Draw  CD  li  to  AB,  cutting  the  circumference  at  D,  and 
from  D  draw  DP  _L  to  AB. 


Now, 


DP'  =  APXPB 


(282) 


.  • .  AP  is  the  base  and  PB  is  the  altitude  of  the  required 
rectangle.  And  APFE,  whose  altitude  PF  =  PB,  is  the 
required  rectangle.  Q.  E.  F. 


ELEMENTS  OF  PLANE  GEOMETRY. 


175 


PROBLEM    XII. 

312.   To  construct  a  rectangle,  given  its  area  and  the  differ- 
ence of  the  base  and  altitude. 

Let  AB  equal  the  given  difference  of  the  base  and  altitude, 
and  let  the  given  area  be  that  of  the  square  whose  side  is  a. 


C 


\ 


A^ 


-\B 


Yd 


On  AB  as  a  diameter,  describe  a  O. 

At  A  draw  the   tangent  AC  =  a,  and   draw  the  secant 
CD  so  as  to  pass  through  the  centre  0. 


Now, 
and 


CD  X  CP=  CA'  =  a\ 
CD  —CP=^PD  =  AB: 


(287) 


.  • .     CD  is  the  base  and  CP  is  the  altitude  of  the  rectangle 
required,  and  the  rectangle  can  readily  be  constructed. 

Q.  E.  F. 


176 


ELEMENTS  OF  PLANE  GEOMETRY. 


PROBLEM    XIII. 

313.   To  construct  a  square  having  a  given  ratio  to  a  given 
square. 

Let  0  :  J)  ho,  the  given    ratio,  and   S  the   given   square 
whose  side  is  a. 


V 

- 

A 

I) 

/                  y"                  1 
/               y                       1 
1            /                             ' 

s 

a 

\ 

1 

F^ 

B 

C 

M 


On  the  indefinite  line  AM,  cut  off  J.^  =  o,  and  BC  =  p. 

On  J.  C  as  a  diameter,  describe  a  semicircle. 

At  B  erect  a  _L  cutting  the  circumference  at  D,  and  draw 
AD  and  CD. 

Take  DF  =  a,  and  draw  EF  II  to  AC. 

.Now,  D^  :  DF  ::   DA  :  i)C  (267),  which  gives 

DE'  :  DT  ::   DA'  :  DC\  (170) 

AB  :  BC  ::   0  :  p;         (283) 


But         DA'  :  DC' 


DF' 


::   0  :  p, 

0  :  p; 


(164) 


DE 

or  DE'  :  a" 

.  • .  the  square  whose  side  is  DE  is  the  required  one.    Q.  E.F. 


314.  Scholium. — Since  similar  polygons  are  to  each  other 
as  the  squares  of  their  homologous  sides,  we  can  find,  by 
means  of  the  above  problem,  the  homologous  side  of  a  poly- 
gon similar  and  having  a  given  ratio  to  a  given  polygon. 


ELEMENTS  OF  PLANE  GEOMETRY.  Ill 


EXEECISES    1^    H^YENTIOK 


THEOREMS. 

1.  The  square  inscribed  in  a  circle  is  equivalent  to  half 
the  square  described  on  the  diameter. 

2.  Prove  geometrically  that  the  square  described  on  the 
sura  of  two  lines  is  equivalent  to  the  sum  of  the  squares 
described  on  the  two  lines  plus  twice  the  rectangle  of  the 
lines. 

3.  Prove  geometrically  that  the  square  described  on  the 
difference  of  two  lines  is  equivalent  to  the  sum  of  the 
squares  described  on  the  two  lines  minus  twice  the  rectangle 
of  the  lines. 

4.  Prove  geometrically  that  the  rectangle  of  the  sum  and 
difference  of  two  lines  is  equivalent  to  the  difference  of  the 
squares  described  on  the  lines. 

5.  If  a  straight  line  is  drawn  from  the  vertex  of  an 
isosceles  triangle  to  any  point  in  the  base,  the  square  of  this 
line  is  equivalent  to  the  rectangle  of  the  segments  of  the 
base  together  with  the  square  of  either  of  the  equal  sides. 

6.  The  area  of  a  circumscribed  polygon  equals  half  the 
product  of  the  perimeter  and  the  radius  of  the  inscribed 
circle. 

7.  The  triangle  formed  by  drawing  straight  lines  from  the 
extremities  of  one  of  the  non-parallel  sides  of  a  trapezoid  to 
the  middle  point  of  the  other,  is  equivalent  to  half  the 
trapezoid. 

8.  The  two  triangles  formed  by  drawing  straight  lines 
from  any  point  within  a  parallelogram  to  the  extremities  of 
either  pair  of  opposite  sides,  are  equivalent  to  half  the 
parallelogram. 


178  ELEMENTS  OF  PLANE  GEOMETRY. 


9.  The  bisector  of  the  vertical  angle  of  a  triangle  divides 
the  base  into  parts  proportional  to  the  adjacent  sides  of  the 
triangle. 


PROBLEMS. 

1.  Trisect  a  given  straight  line. 

2.  Bisect  a  parallelogram  by  a  line  passing  through  any 
given  point  in  the  perimeter. 

3.  Construct  a  parallelogram  whose  surface  and  perimeter 
are  respectively  equal  to  the  surface  and  perimeter  of  a  given 
triangle. 

4.  On  a  given  straight  line  construct  a  rectangle  equivalent 
to  a  given  rectangle. 

5.  Construct  a  polygon  similar  to  a  given  polygon  and 
whose  area  is  in  a  given  ratio  to  that  of  the  given  polygon. 


ELEMENTS  OF  PLANE  GEOMETRY. 


179 


BOOK  V. 

REGULAE   POLYGOI^S  AND   THE 
CIRCLE. 


DEFINITION. 


315.  A  Regular  Polygon  is  one  which  is  both  equilateral 
and  equiangular. 


THEOREM    h 

316.  Any  equilateral  polygon  inscribed  in  a  circle  is  regular. 
Let  F  be  an  equilateral  polygon  inscribed  in  a  O  • 


To  prove  that  P  is  a  regular  polygon. 

Arc  AB  =:  arc  BC  =  arc  CD  =  arc  DE,  etc.;     (189) 

arc  ABC  =  arc  BCD,  etc.;  (Ax.  2.) 

^b  =  Z-c  =  Z.d,  etc.;  (207) 

P  is  a  regular  polygon.     (315)     Q.  E.  D. 


180  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    n. 

317.  A   circle    can    be    circumscribed    about    any  regular 
polygon. 

Let  ABCDEF  be  a  regular  polygon. 


F  ^ 

--.7? 

^\ 

f 

•^      V\ 

\ 

0            ") 

,B^ 

^_--'-^' 

D 


To  prove  that  a  O  can  be  circumscribed  about  ABCDEF. 

Describe  a  circumference  through  the  vertices  A^  F,  and  E. 

(193) 

From  0,  the  centre,  draw  OP  _L  to  FE,  bisecting  it  at  P, 
and  draw  OA  and  OD. 

On  OP  as  an  axis,  revolve  the  quadrilateral  AOPF  till 
it  falls  in  the  plane  of  ODEP. 

Since  OP  is  _L  to  FE,  PF  falls  on  its  equal  PE,  F  billing 
on  E;  and  since  /-.  a  =  Z_  b,  FA  falls  on  its  equal  ED^ 
A  falling  on  D; 

OA  =  OD,  and  the  circumference  passes  through  D. 

Likewise  we  can  prove  that  the  circumference  passing 
through  F,  E,  and  D  passes  also  through  the  vertex  C,  and  thus 
through  all  the  successive  vertices  of  the  polygon.      Q.  E.  D. 

318.  Cor. — A  circle  can  be  inscribed  in  a  regular  polygon. 


ELEMENTS  OF  PLANE  GEOMETRY.  181 


DEFINITIONS. 

319.  The  Centre  of  a  regular  polygon  is  the  common 
centre  of  its  circumscribed  and  inscribed  circles. 

320.  The  Angle  at  tUe  Centre  of  a  regular  polygon  is 
the  angle  formed  by  two  lines  drawn  from  the  centre  to 
the  extremities  of  any  side. 

The  angles  at  the  centre  are  equal,  any  one  being  equal  to  four 
right  angles  divided  by  the  number  of  sides  of  the  polygon. 

321.  The  Apotheni  of  a  regular  polygon  is  the  perpen- 
dicular distance  from  the  centre  to  any  side,  and  is  equal  to 
the  radius  of  the  inscribed  circle. 


16 


182  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM    III. 

322.  Regular  polygons  of  the  same  number  of  sides  are 
similar  figures. 

Let  p  and  P  be  regular  polygons  of  the  same  number 
of  sides. 


E  D  0  N 


A  B  G  H 

To  prove  that  p  and  P  are  similar  figures. 

Since  p  and  P  are  regular  and  have  the  same  number 
of  sides,  they  are  mutually  equiangular.  (114) 

Also  AB  '.  BC  ::   I  \  1, 

and                       GH  :  HM  ::   1   :  1;  (315) 

AB  :  BC  ::    GR  :  HM;  (164) 

p  and  P  are  similar.      (237)  Q.  E.  D. 

323.  Cor.  1. — The  perimeters  of  similar  regular  polygons 
are  to  each  other  as  the  radii  of  their  circumscribed  or  inscribed 
circles  (281). 

324.  Cor.  2. —  Tlie  areas  of  similar  regular  polygons  are  to 
each  other  as  the  squares  of  the  radii  of  their  circumscribed  or 
inscribed  circles  (292). 


ELEMENTS  OF  PLANE  GEOMETRY. 


183 


EELATION  BETWEEIN^  THE  CIBCUM- 

FEREIsrCE    AKD    DIAMETER 

OF    A    CIRCLE. 


THEOREM   IV. 

325.   The  circumferences  of  circles  are  to  each  other  as  their 
radii. 

Let  c  and  C  be  the  circumferences,  r  and  B  the   radii, 
and  d  and  D  the  diameters  of  the  Os  o  and  0. 


To  prove  that  c  :    C  ::   r  :  B. 

Inscribe  in  the  two  circles  similar  regular  polygons,  and 
denote  their  perimeters  by  p  and  P. 


Then 


P 


P  ::  r  :  R. 


(323) 


Now,  this  is  true  whatever  may  be  the  number  of  sides  of 
the  polygons,  if  there  is  the  same  number  in  each ;  hence  it  is 
true  when  the  number  of  sides  is  infinitely  great,  in  which 
case  p  =  c,  and  P  =z  C,  while  r  and  B  remain  the  same; 


B. 


Q.  E.  D. 


184  ELEMENTS  OF  PLANE  GEOMETRY. 

326.  Cor.  1. —  The  circumferences  of  circles  are  to  each  other 
as  their  diameters. 

By  (166)  the  above  proportion  becomes 

c  :    C  ::   2r  :  2B, 

or  c  :   C  ::  d  :  D. 

327.  Cor.  2. —  The  ratio  of  the  circumference  of  a  circle  to 
its  diameter  is  a  constant  quantity. 

By  (160)  the  last  proportion  becomes 

c  :  d  ::    C  :  D, 

or  c    C 

H  ~d' 

This  constant  ratio  is  usually  denoted  by  ;r,  the  Greek 
letter  p,  called  pi.  The  numerical  value  of  t:  can  be  found 
only  approximately,  as  can  be  proved  by  the  higher  mathe- 
matics. 

Hence,  in  any  circle,  the  circumference  and  its  diameter  are 
incommensurable. 

328.  Cor.  3. —  The  ciTcumference  of  a  circle  equals  the 
diameter  multiplied  by  r. 

C 
yr  =  rr,  whence  C  ^=  n  D, 

or  C=27z  B. 


ELEMENTS  OF  PLANE  GEOMETRY. 


185 


THEOREM  V. 

329.  The  area  of  a  regular  polygon  equals  half  the  product 
of  its  perimeter  and  apothem. 

Let  P  be  the  perimeter  and  A  the  apothem  of  the  regular 
polygon  MNORQS. 


To  prove  that  the  area  of  MNORQS  =  i  P  X  ^. 

Draw   CO,   CR,   CQ,  etc.,  dividing   the  polygon   into   as 
many  As  as  it  has  sides. 

All  the  As  have  the  common  altitude  A,  and  the  sum  of 
their  bases  equals  P; 

the  sum  of  the  areas  of  the  As  =  *  P  X  ^,      (247) 

or  the  area  of  the  polygon  =  i  P  X  ^. 

Q.  E.  D. 


16* 


186  ELEMENTS  OF  PLANE  GEOMETRY. 


THEOREM   VI. 

330.   The   area   of  a   circle  equals  half  the  product  of  its 
circumference  and  radius. 

Let  C  be  the  circumference  and  R  the  radius  of  the  O  0. 


To  prove  that  the  area  of  the  circle  =  ^    C  X  i?. 

Inscribe  a  regular  polygon,  and  denote  its  perimeter  by  P, 
and  its  appthem  by  A. 

Then  the  area  of  the  polygon  =  i  F  X  A.  (329) 

Now,  this  is  true  whatever  may  be  the  number  of  sides  of 
the  polygon;  hence  it  is  true  when  the  number  is  infinitely 
great,  in  which  case  P  =   C,  and  A  =  R; 

the  area  of  the  O  =  i  C  X  R-  Q.E.D. 

331.  Cor.  1. —  The  area  of  a  O  ^=  t:  R\  R  heiyig  the 
radius. 

The  area  of  a  O  =  ^  C  X  i^. 

But  O  =  2  t:  R;  (327) 

the  area  of  a  O  =  ??  R^. 

332.  Cor.  2. —  The  area  of  a  sector  of  a  circle  equals  half 
the  product  of  its  arc  and  the  radius. 

DEFINITION". 

333.  Similar  Sectors  are  sectors  of  different  circles, 
which  have  equal  angles  at  the  centre. 


ELEMENTS  OF  PLANE  GEOMETRY.  187 


THEOREM    VII. 

334.   Circles  are  to  each  other  as  the  squares  of  their  radii. 
Let  r  and  E  denote  the  radii  of  the  Os  o  and  0. 


To  prove  that  o  :  0   ::   r^  :  R"^ 


and  0  =  n  R\  (331) 

Divide;  then 

0  TT  r'         r' 

or  0  :  0  ::  r''  :  R\  Q.E.D. 

335.  Cor. — Similar  sectors  are  to  each  other  as  the  squares 
of  their  radii. 


188 


ELEMENTS  OF  PLANE  GEOMETRY, 


PROBLEMS    IN    COI^^STEUCTIOK 


PROBLEM  I. 

336.   To  inscribe  a  square  in  a  given  circle. 
Let  0  be  the  centre  of  the  given  O- 

C 


Draw  any  two  diameters,  as  AB  and  CD,  _L  to  each 
other,  and  draw  A  C,  CB,  BDj  and  AD. 

Now  the  angles  about  the  centre  are  equal  ; 

the  circumference  is  divided  into  four  equal  arcs; 

the  chords  A  0,  CB,  BD,  and  AD  are  equal.     (188) 

The  Z.S  ADB,  DBC,  BCA,  and  CAD  are  Ls;         (208) 

ADBC  is  the  required  square.     (123)     Q.  E.  F. 

337.  Cor. — To  inscribe  a  regular  polygon  of  8  sides,  bisect 
the  arcs  subtended  by  the  sides  of  an  inscribed  square  and  draw 
chords;  and  by  continuing  the  process,  we  can  inscribe  regular 
polygons  of  16,  82,  etc.,  sides. 


ELEMENTS  OF  PLANE  GEOMETRY.  189 

PROBLEM    11. 

338.   To  inscribe  a  regular  hexagon  in  a  circle. 
Let  0  be  the  centre  of  a  given  O. 


Draw  any  radius,  as  OA,  and  with  J.  as  a  centre  and  the 
radius  of  the  circle  describe  an  arc,  cutting  the  circumference 
at  B. 

Draw  AB  and  OB. 

Now,  the  A  AB  0  is  both  equilateral  and  equiangular ;  (97) 

Z.  a  =  i  of  2  Ls  =  i  of  4  Ls;  (81) 

.  • .     arc  AB  =  i  of  the  circumference,  and  the  chord  AB 
is  the  side  of  a  regular  inscribed  hexagon ; 

.  • .     ABCDEF,  which  is  formed  by  applying  the  radius  six 
times  as  a  chord,  is  the  required  hexagon.  Q.  E.  F. 

339.  Cor.  1. —  To  inscribe  an  equilateral  triangle,  join  the 
alternate  vertices  of  a  regular  inscribed  hexagon. 

340.  Cor.  2. —  To  inscribe  a  regular  polygon  of  12  sides, 
bisect  the  arcs  subtended  by  the  sides  of  a  regular  inscribed 
hexagon  and  draw  chords;  and  by  continuing  the  process,  we 
can  inscribe  regular  polygons  of  2J^,  Jf8,  etc.,  sides. 


190  ELEMENTS  OF  PLANE  GEOMETRY, 


PROBLEM    III. 
341.  In  a  given  circle  to  inscribe  a  regular  decagon. 
Let  0  be  the  centre  of  the  given  O . 


Suppose  the  problem  to  be  solved,  and  let  ABC^  etc.,  be 
the  regular  inscribed  decagon. 

Draw  ^C  and  ^i). 

Now,  AC  and  BD  bisect  the  circumference; 

they  are  diameters  and  intersect  at  the  centre  C. 

Draw  BEy  cutting  ^  (7  at  P. 

Z_  a  is  measured  by  h  (arc  AB  +  arc  EC),  or  g  arc  BC, 

(212) 
and  Z^  b  is  measured  by  ^  arc  BC;  (206) 

A  APB  is  isosceles,  and  AB  =  BP, 


ELEMENTS  OF  PLANE  GEOMETRY.  191 


Also,  Z_  c^  is  measured  by  i  arc  ED,  or  arc  AB, 
and         Z_  e  is  measured  by  arc  AB;  ,       (205) 

A  .BOP  is  isosceles,  and  OP  =  BP  =  AB. 
Z_  c  is  measured  by  ^  arc  AE,  or  AB;  (206) 

.  * .  As  APB  and  ABO  are  mutually  equiangular  and  similar; 

(269) 
AO  :  AB  ::  AB  :  AP, 

or  AO  :  OP  ::    OP  :  AP. 

But  this  shows  that  A  0,  the  radius,  is  divided  in  extreme 
and  mean  ratio  at  P,  and  that  OP,  the  greater  part,  equals 
AB,  Si  side  of  the  regular  inscribed  decagon. 

Therefore,  to  inscribe  a  regular  decagon,  divide  the  radius 
in  extreme  and  mean  ratio,  and  apply  the  greater  part  ten 
times  as  a  chord.  Q.  E.  F. 

342.  Cor.  1. —  To  inscribe  a  regular  pentagon,  join  the 
alternate  vertices  of  a  regular  inscribed  decagon. 

343.  Cor. —  To  inscribe  a  regular  polygon  of  20  sides,  bisect 
the  arcs  subtended  by  the  sides  of  a  regular  inscribed  decagon 
and  draw  chords;  and  by  continuing  the  process,  we  can  inscribe 
regular  polygons  of  40,  80,  etc.,  sides. 


192  ELEMENTS  OF  PLANE  GEOMETRY. 


PROBLEM    IV. 

344.  In  a  given  circle  to  inserihe  a  regular  pentadecagon. 
Let  C  be  the  given  O- 


Draw  the  chord  AB  equal  to  the  side  of  a  regular  inscribed 
hexagon,  and  the  chord  BD  equal  to  the  side  of  a  regular 
inscribed  decagon,  and  draw  A  D. 

Now,  arc  AD  =  arc  AB  —  arc  DB 

=  i  of  the  circumference  —  J^  of  the  circumference 

=  y^^  of  the  circumference. 

Therefore  chord  AD  =  a  side  of  a  regular  inscribed  penta- 
decagon;  and  hence  if  we  apply  AD  fifteen  times  as  a  chord 
we  get  the  required  polygon.  Q.  E.  F. 

345.  Cor. — To  inscribe  a  regular  polygon  of  SO  sides, 
bisect  the  arcs  subtended  by  the  sides  of  a  regular  inscribed 
pentadecagon  and  draw  chords;  and  by  continuing  the  process, 
we  can  inscribe  regular  polygons  of  60,  120,  etc.,  sides. 


ELEMENTS  OF  PLANE  GEOMETRY.  193 

PROBLEM    V. 

346.  In  a  circle  whose  radius  is  unity,  to  find  the  value  of 
the  chord  of  half  an  arc  in  terms  of  the  chord  of  the  whole  arc. 

Let  0  be  the  centre  of  a  O  whose  radius  is  1,  AB  the 
chord  of  an  arc,  and  BC  the  chord  of  half  the  arc. 

0 


Draw  the  radii  OB  and  OC. 

In  the         RA  BDO,  OB'  =  OD'  +  BB^. 

Whence  OD  =  Vd^  —  BI)\ 

But         OB  --=00  =  1,  and  BD  =  ^; 


In  the        UaCDB,BC=  V^Wff~-\-  CD\ 

But       CD  =  1  —  02)  -  1  -Vl  —  |-  |- 

Substitute  -^  and  1  —  V  1  —  Hr       ior  then-   equals 


2    '^"^^  ^         '   ^        \  2  / 
BD  and  CD,  and  reduce 


Then  BC  =   y2  —  1/4  —  ^^l  Q.  E.  F. 

17 


194  ELEMENTS  OF  PLANE  GEOMETRY. 

PROBLEM    VI. 

347.   To  find  the  numerical  value  of  iz,  approximately. 
Let  C  be  the  circumference,  and  R  the  radius  of  a  O. 


When  B  =  1 


(327) 


Now,  by  means  of  the  formula  BC  =  \/2  —  VT--A^\ 
established  in  (346),  we  make  the  following  computations : 

In  a  regular  inscribed  polygon  of 
No.  Sides.  Form  of  Computation.         Length  of  Side.     Perimeter. 

6.  See  (338)  1.00000000  6.00000000. 

12.  jB 0=1/2—1/4       —     r        =  .51763809  6.21165708. 

24.  BC=  \/2--/4— (.517638^^==  .26105238  6.26525722. 

48.  5C=l/2— 1/4^261052387^=  .13080626  6.27870041. 

96. 50=1/2—1/4— (.13080626)'^=  .06543817  6.28206396. 

192.  50=  V^2— 1/4— (.065"43817)^=  .03272346  6.28290510. 

384.  50=  1^2— 1/4— (.03272346^=  .01636228  6.28311544. 

768.  50=  l/2-l/4^=(jai6362l8y2=  .00818121  6.28316941. 

It  will  be  seen  that  the  first  four  decimal  places  remain 
the  same,  to  whatever  extent  we  increase  the  number  of  sides. 
Hence  we  can  consider  6.28317  as  the  approximate  value  of 
the  circumference  of  a  circle  whose  radius  is  1 . 

^  ^  O       6^2^  ^  g^^^g  ^^^^,jy  Q  ^  -p 


ELEMENTS  OF  PLANE  GEOMETRY.  195 


EXEECISES    IIsT    il^YEl^TIOl^. 


THEOREMS. 

1.  The  side  of  an  inscribed  equilateral  triangle  equals  half 
the  side  of  the  circumscribed  equilateral  triangle. 

2.  The  diameter  of  a  circle  is  a  mean  proportional  between 
the  sides  of  the  equilateral  triangle  and  the  regular  hexagon 
circumscribed  about  the  circle. 

3.  The  square  inscribed  in  a  circle  equals  half  the  square 
on  the  diameter. 

4.  The  area  of  a  regular  inscribed  hexagon  equals  three- 
fourths  the  area  of  a  regular  circumscribed  hexagon. 

5.  The  area  of  a  regular  inscribed  hexagon  is  a  mean 
proportional  between  the  areas  of  the  inscribed  and  circum- 
scribed equilateral  triangles. 

6.  If  the  vertices  of  a  square  are  taken  as  centres  and  half 
the  diagonal  as  a  radius  and  circles  be  described,  the  points 
of  intersection  of  the  circumferences  and  the  sides  of  the 
square  are  the  vertices  of  a  regular  octagon. 

7.  The  area  of  a  regular  inscribed  octagon  equals  the  area 
of  a  rectangle  whose  adjacent  sides  equal  the  sides  of  the 
circumscribed  and  inscribed  squares. 

8.  The  area  of  a  regular  inscribed  dodecagon  equals  three 
times  the  square  on  the  radius. 


196  ELEMENTS  OF  PLANE  GEOMETRY, 


PROBLEMS. 

1.  Inscribe  in  a  given  circle  a  regular  polygon  similar  to  a 
given  regular  polygon. 

2.  Circumscribe  a  polygon  similar  to  a  given  inscribed 
polygon.     , 

3.  In  a  given  circle,  inscribe  three  equal  circles,  touching 
each  other  and  the  given  circle. 

4.  In  a  given  circle,  inscribe  four  equal  circles  in  mutual 
contact  with  each  other  and  the  given  circle. 

5.  In  a  given  equilateral  triangle,  inscribe  three  equal 
circles,  touching  each  other,  and  each  touching  two  sides  of 
the  triangle. 

6.  About  a  given  circle,  describe  six  circles,  each  equal  to 
the  given  one  and  in  mutual  contact  with  each  other  and  the 
given  circle. 


O 


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